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course Phy 121
9/16 10:50 pm
006. Using equations with uniformly accelerated motion.
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Question: `q001. Note that there are 12 questions in this assignment.
Using the equation vf = v0 + a * `dt determine the acceleration of an object whose velocity increases at a uniform rate from 10 m/s to 30 m/s in 15 seconds. Begin by solving the equation for the acceleration a, then 'plug in' your initial and final velocities. Describe your work step y step.
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Your solution:
vf = v0 + a * `dt First get the v0 to the other side by subtracting v) from both sides
vf-v0 = a * `dt Now we divide both sides by `dt to isolate a
(vf - v0)/`dt =a Switch sides
a = (vf - v0)/`dt Plug in the values we know
a = (30m/s - 10m/s)/15 s
a = (20m/s)/15s
a = approx. 1.3m/s^2
confidence rating #$&*:
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Given Solution:
The equation vf = v0 + a * `dt is solved for a by first adding -v0 to both sides to obtain vf - v0 = v0 + a * `dt - v0, which simplifies to vf - v0 = a * `dt. Both sides are then divided by `dt to obtain (vf - v0) / `dt = a. Reversing left-and right-hand sides we obtain the formula a = (vf - v0) / `dt.
We then plug in our given values of initial and final velocities and the time interval. Since velocity increases from 10 m/s to 30 m/s, initial velocity is v0 = 10 m/s and final velocity is vf = 30 m/s. The time interval `dt is 15 seconds, so we have a = (30 m/s - 10 m/s) / (15 s) = 20 m/s / (15 s) = 1.33.. m/s^2.
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Self-critique (if necessary):Ok
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Self-critique rating:OK
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Question: `q002. It wasn't necessary to use a equation to solve this problem. How could this problem had been reasoned out without the use of an equation?
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Your solution:
We can reason that the acceleration can be determined by finding the change in velocity and dividing it by the time interval. The change in velocity is 30m/s - 10m/s which is 20m/s. 20m/s divided by the time interval of 15 seconds is 1.3m/s/s or 1.3 m/s^2.
confidence rating #$&*:
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Given Solution:
Knowing that acceleration is the average rate at which velocity changes, we would first find the change in velocity from 10 meters/second to 30 meters/second, which is 20 meters/second. We would then divided change in velocity by the time interval to get 20 meters/second / (15 sec) = 1.33 m/s^2.
STUDENT QUESTION (about reasoning vs. using the equation)
I understand but the steps taken to get to the acceleration were the steps of the equation?????
INSTRUCTOR RESPONSE
The steps outlined here are the steps we could use to derive the equation. However it's possible to use the equation blindly, without understanding the reasoning behind it. In fact this is how most student use the equation, if not asked questions of this nature about the reasoning.
So, this question asks for the reasoning.
The first statement in the given solution is
'Knowing that acceleration is the average rate at which velocity changes, we would first find the change in velocity from 10 meters/second to 30 meters/second, which is 20 meters/second.'
When using the equation you never explicitly find or reason out the change in velocity, though of course the change in velocity is there in the equation, represented by the term a * `dt. In other words, you do find it, but you can use the equation without ever recognizing that you have done so.
Similarly the step a = (30 m/s - 10 m/s) / 15 s in your equation-based solution does correctly divide the change in velocity by the time interval, but you can use the equation to do this without ever recognizing that you have done so.
The direct reasoning solution never mentions or uses the equation, though of course direct reasoning can be used to derive the equation.
This should help illustrate the difference between direct reasoning and using an equation. Both skills are important.
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Self-critique (if necessary):OK
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Self-critique rating:OK
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Question: `q003. Use the equation `ds = (vf + v0) / 2 * `dt to determine the initial velocity of an object which accelerates uniformly through a distance of 80 meters in 10 seconds, ending up at a velocity of 6 meters / sec. begin by solving the equation for the desired quantity. Show every step of your solution.
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Your solution:
`ds =(vf+v0)/2 *`dt First multiply each side by 1/`dt
`ds/`dt = =(vf+v0)/2 Now we multiply each side by 2
2 * `ds/`dt = vf+v0 Subtract vf from each side
2* `ds/`dt - vf= v0 Reverse the sides
v0 = 2* `ds/`dt -vf Plug in our values
v0 = 2* (80m/10s) -6m/s Simplify
v0 = 2 *8m/s -6m/s Simplify
v0 = 16m/s - 6m/s Simplify
v0 = 10m/s
confidence rating #$&*:
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Given Solution:
We begin by solving the equation for v0. Starting with
`ds = (vf + v0) / 2 * `dt, we can first multiply both sides of the equation by 2 / `dt, which gives us
`ds * 2 / `dt = (vf + v0) / 2 * `dt * 2 / `dt. The right-hand side can be rearranged to give
(vf + v0) * `dt / `dt * 2 / 2; since `dt / `dt = 1 and 2 / 2 = 1
the right-hand side becomes just vf + v0. The equation therefore becomes
2 * `ds / `dt = vf + v0. Adding -vf to both sides we obtain
v0 = 2 * `ds / `dt - vf.
We now plug in `ds = 80 meters, `dt = 10 sec and vf = 6 m/s to get
v0 = 2 * 80 meters / 10 sec - 6 m/s = 160 meters / 10 sec - 6 m/s = 16 m/s - 6 m/s = 10 m/s.
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Self-critique (if necessary):OK
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Self-critique rating:OK
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Question: `q004. We can reconcile the above solution with straightforward reasoning. How could the initial velocity have been reasoned out from the given information without the use of an equation? Hint: two of the quantities given in the problem can be combined to give another important quantity, which can then be combined with the third given quantity to reason out the final velocity.
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Your solution:
We know the object traveled 80 meters in 10 seconds so its average velocity was 80m/10s or 8m/s. We know the final velocity was 6m/seconds. If the average velocity is 8m/s and the final velocity is 6m/s its easy to see that the initial velocity would be 10m/s, but just to be sure we can do some math. Since the average velocity is the sum of the initial and final velocities divided by two then the sum must be 16m/s. 16m/s - 6m/s = 10m/s which is our initial velocity.
confidence rating #$&*:
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Given Solution:
The average velocity of the object is the average rate at which its position changes, which is equal to the 80 meters change in position divided by the 10 s change in clock time, or 80 meters / 10 sec = 8 meters / sec. Since the 8 m/s average velocity is equal to the average of the unknown initial velocity and the 6 m/s final velocity, we ask what quantity when average with 6 m/s will give us 8 m/s. Knowing that the average must be halfway between the two numbers being averaged, we see that the initial velocity must be 10 m/s. That is, 8 m/s is halfway between 6 m/s and 10 m/s.
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Self-critique (if necessary):OK
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Self-critique rating:OK
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Question: `q005. Using the equation `ds = v0 `dt + .5 a `dt^2 determine the initial velocity of an object which accelerates uniformly at -2 m/s^2, starting at some unknown velocity, and is displaced 80 meters in 10 seconds. Begin by solving the equation for the unknown quantity and show every step.
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Your solution:
`ds = v0 `dt + .5 a `dt^2 Subtract .5 a `dt^2 from each sided
`ds -.5 a `dt^2 = v0`dt Multiply each side by 1/`dt
(`ds -.5 a `dt^2)/`dt = v0 Plug in our values
(80m- .5*-2m/s^2 *10s^2)10s = v0 Simplify
(80m +1m/s^2*100s^2)/10s =v0 Simplify
(80m + 100m)/10s =v0
180m/10s = v0
18m/s =v0
confidence rating #$&*:
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Given Solution:
The unknown quantity is the initial velocity v0. To solve for v0 we start with
`ds = v0 `dt + .5 a `dt^2. We first add -.5 a `dt^2 to both sides to obtain
`ds - .5 a `dt^2 = v0 `dt. We then divide both sides by `dt to obtain
(`ds - .5 a `dt^2) / `dt = v0.
Then we substitute the given displacement `ds = 80 meters, acceleration a = -2 m/s^2 and time interval `dt = 10 seconds to obtain
v0 = [ 80 meters - .5 * (-2 m/s^2) * (10 sec)^2 ] / (10 sec)
= [ 80 meters - .5 * (-2 m/s^2) * 100 s^2 ] / (10 sec)
= [ 80 m - (-100 m) ] / (10 sec)
= 180 m / (10 s) = 18 m/s.
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Self-critique (if necessary):Ok
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Self-critique rating:Ok
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Question: `q006. Check the consistency of this result by verifying, by direct reasoning rather than equations, that an object whose initial velocity is 18 m/s and which accelerates for 10 seconds at an acceleration of -2 m/s^2 does indeed experience a displacement of 80 meters.
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Your solution:
If the object starts at 18m/s and decreases in velocity 2m/s for every second, over the course of 10 seconds its velocity would change -20m/s and thus its final velocity would be -2m/s. Its average velocity would be 18m/s + 2m/s divided by two or (16m/s)/2 which equals 8m/s. Travelling at an average velocity of 8m/s for 10 seconds the object would travel 80 meters. It checks out.
confidence rating #$&*:
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Given Solution:
The change in the velocity of the object will be -2 m/s^2 * 10 s = -20 m/s.
The object will therefore have a final velocity of 18 m/s - 20 m/s = -2 m/s.
Its average velocity will be the average (18 m/s + (-2 m/s) ) / 2 = 8 m/s.
An object which travels at an average velocity of 8 m/s for 10 sec will travel 80 meters.
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Self-critique (if necessary):OK
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Self-critique rating:OK
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Question: `q007. Using the equation vf^2 = v0^2 + 2 a `ds determine the initial velocity of an object which attains a final velocity of 20 meters/second after accelerating uniformly at 2 meters/second^2 through a displacement of 80 meters. Begin by solving the equation for the unknown quantity and show every step.
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Your solution:
We need to solve for v0 -our only unknown.
vf^2 = v0^2 + 2 a `ds Subtract 2a`ds from both sides
vf^2-2a`ds = v0^2 take the square root of each side
sqrt(vf^2-2a`ds)=sqrt(v0^2) Reverse sides
v0^2 = sqrt(vf^2-2a`ds)
v0^2 = sqrt [(20m/s)^2 -2(2m/s^2)80m]
v0^2 = sqrt(400m^2/s2 -320m^2/s^2)
v0^2 = sqrt(80m^2/s^2)
v0^2 = approx. 8.9m/s
confidence rating #$&*:
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Given Solution:
To solve for the unknown initial velocity v0 we start with
vf^2 = v0^2 + 2 a `ds. We first add -2 a `ds to both sides to obtain
vf^2 - 2 a `ds = v0^2. We then reverse the right-and left-hand sides and take the square root of both sides, obtaining
v0 = +- `sqrt( vf^2 - 2 a `ds).
We then substitute the given quantities vf = 20 m/s, `ds = 80 m and a = 3 m/s^2 to obtain
v0 = +- `sqrt( (20 m/s)^2 - 2 * 2 m/s^2 * 80 m) = +- `sqrt( 400 m^2 / s^2 - 320 m^2 / s^2) = +- `sqrt(80 m^2 / s^2) = +- 8.9 m/s (approx.).
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Self-critique (if necessary):OK
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Self-critique rating:OK
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Question: `q008. We can verify that starting at +8.9 m/s an object which attains a final velocity of 20 m/s while displacing 80 meters must accelerate at 2 m/s^2. In this case the average velocity will be ( 8.9 m/s + 20 m/s) / 2 = 14.5 m/s (approx) and the change in velocity will be 20 m/s - 8.9 m/s = 11.1 m/s.
At average velocity 14.5 meters/second the time required to displace the 80 meters will be 80 m / (14.5 sec) = 5.5 sec (approx).
The velocity change of 11.1 meters/second in 5.5 sec implies an average acceleration of 11.1 m/s / (5.5 sec) = 2 m/s^2 (approx), consistent with our results.
Verify that starting at -8.9 m/s the object will also have acceleration 2 meters/second^2 if it ends up at velocity 20 m/s while displacing 80 meters.
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Your solution:
At first this seems impossible that it would be the case but raising things to a power of two can do interesting things.
We can plug our values into the previous equation and see if indeed both sides are equal
vf^2 = v0^2 + 2 a `ds
(20m/s)^2 = (-8.9m/s)^2 + 2(2m/s^2)80m
400m^2/s^2 =80m^2/s^2 + 4m/s*80m
400m^2/s^2 =80m^2/s^2 + 320m^2/^2
400m^2/s^2 = 400m^2/s^2
confidence rating #$&*:
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Given Solution:
In this case the average velocity will be ( -8.9 m/s + 20 m/s) / 2 = 5.5 m/s (approx) and the change in velocity will be 20 m/s - (-8.9 m/s) = 28.9 m/s (approx). At average velocity 5.5 meters/second the time required to displace the 80 meters will be 80 m / (5.5 sec) = 14.5 sec (approx). The velocity change of 28.5 meters/second in 14.5 sec implies an average acceleration of 28.5 m/s / (14.5 sec) = 2 m/s^2 (approx), again consistent with our results.
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Self-critique (if necessary):I verified this a different way, but I grasp the concept in your solution.
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Self-critique rating:OK
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Question: `q009. Describe in commonsense terms the motion of the object in this example if its initial velocity is indeed -8.9 m/s. Assume that the object starts at the crossroads between two roads running North and South, and East and West, respectively, and that the object ends up 80 meters North of the crossroads. In what direction does it start out, what happens to its speed, and how does it end up where it does?
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Your solution:
The object is heading South at first at 8.9m/s. It slows down at 2 m/s^2 until it reaches zero m/s and then reverses direction and heads north, picking up speed at 2m^2 and continues on eventually ending up 80 meters North of the crossroads.
confidence rating #$&*:
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Given Solution:
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Question: `q010. An object speeds up from 10 m/s to 20 m/s, accelerating uniformly and traveling 60 meters during this interval.
Specify which of the quantities v_0, v_f, aAve, `ds and `dt are given, and specify the value of each.
Specify which of the four equations of uniformly accelerated motion include the given three quantities. There is at least one such equation, and there might be two.
For each of the equations you specified, identify the quantity for which the value is not given. Then symbolically solve the equation for each of these quantities, showing the steps of your algebra.
Substitute the three given quantities into your solution, and simplify.
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Your solution:
v0 = 10m/s
vf = 20m/s
`ds = 60m
`ds = (v0 + vf) / 2 * `dt Multiply each side by 2/(v0+vf)
2`ds/(v0+vf)=`dt Substitute
2* 60 m/(10m/s +20m/s) = `dt
120m/(30m/s) = `dt Reverse
`dt = 120m/(30m/s) simplify
`dt = 40s
@&
120/30 = 4, not 40
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vf^2 = v0^2 + 2 a `ds add - 2a`ds to each side
vf^2- 2a`ds = v0^2 add -vf^2 to each side
-2a`ds =v0^2 -vf^2 divide each side by -2`ds
a = (v0^2 -vf^2)/-2`ds Substitute
a = [(10m/s)^2 - (20m/s)^2)]/-2*60m
a= (100m^2/s^2 -400m^2/s^2)/-120m
a= (-300m^2/S^2)/-120m
a = 2.5m/s^2
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Very good.
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confidence rating #$&*:
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Question: `q011. An cart initially moving at 10 cm/s travels 40 cm while accelerating at 5 m/s^2. Using the equations of uniformly accelerated motion determine the time required and the the cart's final velocity.
Hint: You will need to start out with either the third or the fourth equation of uniformly accelerated motion. You are advised that it's easier to start out with the fourth equation.
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Your solution:
vf^2 = v0^2 + 2 a `ds substitute (and convert 5 meters to 500cm)
= (10cm/s)^2+500cm/s^2*40cm
=100cm^2/s^2 +20000cm^2/s^2
vf^2 =20100cm^2/m^2 Take sqrt of each side
vf = approx 142cm/s
vf = v0 + a * `dt solve for `dt add -v0 to each side
vf-v0 = a*`dt divide each side by a
(vf-v0)/a=`dt reverse
`dt =(vf-v0)/a substitute
`dt = (142cm/s-10cm/s)/500cm/s^2
`dt = (132cm/s)/ 500cm/s^2
`dt = .264sec
confidence rating #$&*:
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Question: `q012. An object starts at position x = +10 cm with a velocity of +5 cm/s, and accelerates uniformly, ending up at position x = -30 cm after a time interval of 8 seconds. What is its velocity at this point, and what was its acceleration during this interval?
Principles of Physics students should not spend over 5 minutes on this problem, General College Physics students should not spend over 10 minutes.
University Physics students are expected to be able to solve this problem, but if it hasn't been solved within 15 minutes, should submit their best thinking and await the instructor's notes.
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Your solution:
v0=+5cm/s
`dt=8s
`ds =40cm (or is it -40cm?) I値l be able to figure it out from the answers I get.
@&
From x = +10 cm to x = -30 cm the displacement is `dx = -30 cm - 10 cm = -40 cm.
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Velocity likely to be negative as is acceleration.
`ds = (v0 + vf) / 2 * `dt Solve for vf - divide both sides by 2*`dt
`ds/2*`dt = v0 +vf add -v0 to each side
`ds/2*`dt -v0 = vf reverse
vf = `ds/2*`dt -v0 substitute
vf = -40cm/(2*8s) -5cm/s
vf = -40cm(16s) - 5cm/s
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-40 cm / (16 s) = -2.5 cm/s. So vf will be -7.5 cm/s, not -2.5 cm/s.
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vf = 2.5cm/s - 5cm/s
vf = -2.5cm/s
I would move on to acceleration from here but I知 already 7 minutes in and that 2 minutes more than I should do.
@&
Except for that sign error you've pretty much nailed it.
Change in velocity from + 5 cm/s to -7.5 cm/s is -12.5 cm/s, which takes place in 8 seconds, giving you an acceleration of -1.56 cm/s^2 (rounded to probably one too many significant figures)
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confidence rating #$&*:
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The object ends up at position +80 meters, which is assumed to be 80 meters to the North. Its initial velocity is -8.9 m/s, with the - sign indicating that the initial velocity is in the direction opposite to the displacement of the object. So the object must start out moving to the South at 8.9 meters/second.
Its acceleration is +2 m/s^2, which is in the opposite direction to its initial velocity. This means that the velocity of the object changes by +2 m/s every second. After 1 second the velocity of the object will therefore be -8.9 m/s + 2 m/s = -6.9 m/s. After another second the velocity will be -6.9 m/s + 2 m/s = -4.9 m/s. After another second the velocity will be -2.9 m/s, after another -.9 m/s, and after another -.9 m/s + 2 m/s = +1.1 m/s. The speed of the object must therefore decrease, starting at 8.9 m/s (remember speed is always positive because speed doesn't have direction) and decreasing to 6.9 m/s, then 4.9 m/s, etc. until it reaches 0 for an instant, and then starts increasing again.
Since velocities after that instant become positive, the object will therefore start moving to the North immediately after coming to a stop, picking up speed at 2 m/s every second. This will continue until the object has attained a velocity of +20 meters/second and has displaced +80 meters from its initial position.{}{}It is important to understand that it is possible for velocity to be in one direction and acceleration in the other. In this case the initial velocity is negative while the acceleration is positive. If this continues long enough the velocity will reach zero, then will become positive.
STUDENT QUESTION
I understood the negative velocity but was unsure how to explain the rest. I am still rather confused by the last paragraph, expecially where it says that it is possible for velocity to be in one direction and acceleration in the other.
INSTRUCTOR RESPONSE
If you speed up the acceleration is in the direction of motion.
If you slow down the acceleration is opposite the direction of motion.
To speed up a wagon you can get behind it and push in the direction of its motion, giving it an acceleration in its direction of motion.
To slow it down you can get in front of it and push it against its direction of motion (not advisable if it's a big wagon; think of stopping a child in a small wagon), giving it an acceleration in the direction opposite its motion.
STUDENT COMMENTS
Made relative sense, but still unsure in doubt to my answer that the object was increasing after it was moving in the
right direction. Being negative, would it have started towards the south at that acceleration, and moving north would it have
diminished its negativity? It seems this naturally.
INSTRUCTOR RESPONSE
Accelerating to the north (not 'moving' to the north; it does end up moving to the north, but the object starts out moving to the south), the speed in the southward direction would have diminished, as you say.
Eventually it comes to rest, just for an instant, somewhere south of its starting point. Then the northward acceleration will give it an increasing northward velocity.
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Self-critique (if necessary):OK
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Self-critique rating:OK
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Question: `q009. Describe in commonsense terms the motion of the object in this example if its initial velocity is indeed -8.9 m/s. Assume that the object starts at the crossroads between two roads running North and South, and East and West, respectively, and that the object ends up 80 meters North of the crossroads. In what direction does it start out, what happens to its speed, and how does it end up where it does?
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Your solution:
The object is heading South at first at 8.9m/s. It slows down at 2 m/s^2 until it reaches zero m/s and then reverses direction and heads north, picking up speed at 2m^2 and continues on eventually ending up 80 meters North of the crossroads.
confidence rating #$&*:
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Given Solution:
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Question: `q010. An object speeds up from 10 m/s to 20 m/s, accelerating uniformly and traveling 60 meters during this interval.
Specify which of the quantities v_0, v_f, aAve, `ds and `dt are given, and specify the value of each.
Specify which of the four equations of uniformly accelerated motion include the given three quantities. There is at least one such equation, and there might be two.
For each of the equations you specified, identify the quantity for which the value is not given. Then symbolically solve the equation for each of these quantities, showing the steps of your algebra.
Substitute the three given quantities into your solution, and simplify.
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Your solution:
v0 = 10m/s
vf = 20m/s
`ds = 60m
`ds = (v0 + vf) / 2 * `dt Multiply each side by 2/(v0+vf)
2`ds/(v0+vf)=`dt Substitute
2* 60 m/(10m/s +20m/s) = `dt
120m/(30m/s) = `dt Reverse
`dt = 120m/(30m/s) simplify
`dt = 40s
@&
120/30 = 4, not 40
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vf^2 = v0^2 + 2 a `ds add - 2a`ds to each side
vf^2- 2a`ds = v0^2 add -vf^2 to each side
-2a`ds =v0^2 -vf^2 divide each side by -2`ds
a = (v0^2 -vf^2)/-2`ds Substitute
a = [(10m/s)^2 - (20m/s)^2)]/-2*60m
a= (100m^2/s^2 -400m^2/s^2)/-120m
a= (-300m^2/S^2)/-120m
a = 2.5m/s^2
@&
Very good.
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confidence rating #$&*:
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Question: `q011. An cart initially moving at 10 cm/s travels 40 cm while accelerating at 5 m/s^2. Using the equations of uniformly accelerated motion determine the time required and the the cart's final velocity.
Hint: You will need to start out with either the third or the fourth equation of uniformly accelerated motion. You are advised that it's easier to start out with the fourth equation.
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Your solution:
vf^2 = v0^2 + 2 a `ds substitute (and convert 5 meters to 500cm)
= (10cm/s)^2+500cm/s^2*40cm
=100cm^2/s^2 +20000cm^2/s^2
vf^2 =20100cm^2/m^2 Take sqrt of each side
vf = approx 142cm/s
vf = v0 + a * `dt solve for `dt add -v0 to each side
vf-v0 = a*`dt divide each side by a
(vf-v0)/a=`dt reverse
`dt =(vf-v0)/a substitute
`dt = (142cm/s-10cm/s)/500cm/s^2
`dt = (132cm/s)/ 500cm/s^2
`dt = .264sec
confidence rating #$&*:
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Question: `q012. An object starts at position x = +10 cm with a velocity of +5 cm/s, and accelerates uniformly, ending up at position x = -30 cm after a time interval of 8 seconds. What is its velocity at this point, and what was its acceleration during this interval?
Principles of Physics students should not spend over 5 minutes on this problem, General College Physics students should not spend over 10 minutes.
University Physics students are expected to be able to solve this problem, but if it hasn't been solved within 15 minutes, should submit their best thinking and await the instructor's notes.
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Your solution:
v0=+5cm/s
`dt=8s
`ds =40cm (or is it -40cm?) I値l be able to figure it out from the answers I get.
@&
From x = +10 cm to x = -30 cm the displacement is `dx = -30 cm - 10 cm = -40 cm.
*@
Velocity likely to be negative as is acceleration.
`ds = (v0 + vf) / 2 * `dt Solve for vf - divide both sides by 2*`dt
`ds/2*`dt = v0 +vf add -v0 to each side
`ds/2*`dt -v0 = vf reverse
vf = `ds/2*`dt -v0 substitute
vf = -40cm/(2*8s) -5cm/s
vf = -40cm(16s) - 5cm/s
@&
-40 cm / (16 s) = -2.5 cm/s. So vf will be -7.5 cm/s, not -2.5 cm/s.
*@
vf = 2.5cm/s - 5cm/s
vf = -2.5cm/s
I would move on to acceleration from here but I知 already 7 minutes in and that 2 minutes more than I should do.
@&
Except for that sign error you've pretty much nailed it.
Change in velocity from + 5 cm/s to -7.5 cm/s is -12.5 cm/s, which takes place in 8 seconds, giving you an acceleration of -1.56 cm/s^2 (rounded to probably one too many significant figures)
*@
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The object ends up at position +80 meters, which is assumed to be 80 meters to the North. Its initial velocity is -8.9 m/s, with the - sign indicating that the initial velocity is in the direction opposite to the displacement of the object. So the object must start out moving to the South at 8.9 meters/second.
Its acceleration is +2 m/s^2, which is in the opposite direction to its initial velocity. This means that the velocity of the object changes by +2 m/s every second. After 1 second the velocity of the object will therefore be -8.9 m/s + 2 m/s = -6.9 m/s. After another second the velocity will be -6.9 m/s + 2 m/s = -4.9 m/s. After another second the velocity will be -2.9 m/s, after another -.9 m/s, and after another -.9 m/s + 2 m/s = +1.1 m/s. The speed of the object must therefore decrease, starting at 8.9 m/s (remember speed is always positive because speed doesn't have direction) and decreasing to 6.9 m/s, then 4.9 m/s, etc. until it reaches 0 for an instant, and then starts increasing again.
Since velocities after that instant become positive, the object will therefore start moving to the North immediately after coming to a stop, picking up speed at 2 m/s every second. This will continue until the object has attained a velocity of +20 meters/second and has displaced +80 meters from its initial position.{}{}It is important to understand that it is possible for velocity to be in one direction and acceleration in the other. In this case the initial velocity is negative while the acceleration is positive. If this continues long enough the velocity will reach zero, then will become positive.
STUDENT QUESTION
I understood the negative velocity but was unsure how to explain the rest. I am still rather confused by the last paragraph, expecially where it says that it is possible for velocity to be in one direction and acceleration in the other.
INSTRUCTOR RESPONSE
If you speed up the acceleration is in the direction of motion.
If you slow down the acceleration is opposite the direction of motion.
To speed up a wagon you can get behind it and push in the direction of its motion, giving it an acceleration in its direction of motion.
To slow it down you can get in front of it and push it against its direction of motion (not advisable if it's a big wagon; think of stopping a child in a small wagon), giving it an acceleration in the direction opposite its motion.
STUDENT COMMENTS
Made relative sense, but still unsure in doubt to my answer that the object was increasing after it was moving in the
right direction. Being negative, would it have started towards the south at that acceleration, and moving north would it have
diminished its negativity? It seems this naturally.
INSTRUCTOR RESPONSE
Accelerating to the north (not 'moving' to the north; it does end up moving to the north, but the object starts out moving to the south), the speed in the southward direction would have diminished, as you say.
Eventually it comes to rest, just for an instant, somewhere south of its starting point. Then the northward acceleration will give it an increasing northward velocity.
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Self-critique (if necessary):OK
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Self-critique rating:OK
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Question: `q009. Describe in commonsense terms the motion of the object in this example if its initial velocity is indeed -8.9 m/s. Assume that the object starts at the crossroads between two roads running North and South, and East and West, respectively, and that the object ends up 80 meters North of the crossroads. In what direction does it start out, what happens to its speed, and how does it end up where it does?
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Your solution:
The object is heading South at first at 8.9m/s. It slows down at 2 m/s^2 until it reaches zero m/s and then reverses direction and heads north, picking up speed at 2m^2 and continues on eventually ending up 80 meters North of the crossroads.
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Given Solution:
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Question: `q010. An object speeds up from 10 m/s to 20 m/s, accelerating uniformly and traveling 60 meters during this interval.
Specify which of the quantities v_0, v_f, aAve, `ds and `dt are given, and specify the value of each.
Specify which of the four equations of uniformly accelerated motion include the given three quantities. There is at least one such equation, and there might be two.
For each of the equations you specified, identify the quantity for which the value is not given. Then symbolically solve the equation for each of these quantities, showing the steps of your algebra.
Substitute the three given quantities into your solution, and simplify.
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Your solution:
v0 = 10m/s
vf = 20m/s
`ds = 60m
`ds = (v0 + vf) / 2 * `dt Multiply each side by 2/(v0+vf)
2`ds/(v0+vf)=`dt Substitute
2* 60 m/(10m/s +20m/s) = `dt
120m/(30m/s) = `dt Reverse
`dt = 120m/(30m/s) simplify
`dt = 40s
@&
120/30 = 4, not 40
*@
vf^2 = v0^2 + 2 a `ds add - 2a`ds to each side
vf^2- 2a`ds = v0^2 add -vf^2 to each side
-2a`ds =v0^2 -vf^2 divide each side by -2`ds
a = (v0^2 -vf^2)/-2`ds Substitute
a = [(10m/s)^2 - (20m/s)^2)]/-2*60m
a= (100m^2/s^2 -400m^2/s^2)/-120m
a= (-300m^2/S^2)/-120m
a = 2.5m/s^2
@&
Very good.
*@
confidence rating #$&*:
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Question: `q011. An cart initially moving at 10 cm/s travels 40 cm while accelerating at 5 m/s^2. Using the equations of uniformly accelerated motion determine the time required and the the cart's final velocity.
Hint: You will need to start out with either the third or the fourth equation of uniformly accelerated motion. You are advised that it's easier to start out with the fourth equation.
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Your solution:
vf^2 = v0^2 + 2 a `ds substitute (and convert 5 meters to 500cm)
= (10cm/s)^2+500cm/s^2*40cm
=100cm^2/s^2 +20000cm^2/s^2
vf^2 =20100cm^2/m^2 Take sqrt of each side
vf = approx 142cm/s
vf = v0 + a * `dt solve for `dt add -v0 to each side
vf-v0 = a*`dt divide each side by a
(vf-v0)/a=`dt reverse
`dt =(vf-v0)/a substitute
`dt = (142cm/s-10cm/s)/500cm/s^2
`dt = (132cm/s)/ 500cm/s^2
`dt = .264sec
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Question: `q012. An object starts at position x = +10 cm with a velocity of +5 cm/s, and accelerates uniformly, ending up at position x = -30 cm after a time interval of 8 seconds. What is its velocity at this point, and what was its acceleration during this interval?
Principles of Physics students should not spend over 5 minutes on this problem, General College Physics students should not spend over 10 minutes.
University Physics students are expected to be able to solve this problem, but if it hasn't been solved within 15 minutes, should submit their best thinking and await the instructor's notes.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
v0=+5cm/s
`dt=8s
`ds =40cm (or is it -40cm?) I値l be able to figure it out from the answers I get.
@&
From x = +10 cm to x = -30 cm the displacement is `dx = -30 cm - 10 cm = -40 cm.
*@
Velocity likely to be negative as is acceleration.
`ds = (v0 + vf) / 2 * `dt Solve for vf - divide both sides by 2*`dt
`ds/2*`dt = v0 +vf add -v0 to each side
`ds/2*`dt -v0 = vf reverse
vf = `ds/2*`dt -v0 substitute
vf = -40cm/(2*8s) -5cm/s
vf = -40cm(16s) - 5cm/s
@&
-40 cm / (16 s) = -2.5 cm/s. So vf will be -7.5 cm/s, not -2.5 cm/s.
*@
vf = 2.5cm/s - 5cm/s
vf = -2.5cm/s
I would move on to acceleration from here but I知 already 7 minutes in and that 2 minutes more than I should do.
@&
Except for that sign error you've pretty much nailed it.
Change in velocity from + 5 cm/s to -7.5 cm/s is -12.5 cm/s, which takes place in 8 seconds, giving you an acceleration of -1.56 cm/s^2 (rounded to probably one too many significant figures)
*@
confidence rating #$&*:
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The object ends up at position +80 meters, which is assumed to be 80 meters to the North. Its initial velocity is -8.9 m/s, with the - sign indicating that the initial velocity is in the direction opposite to the displacement of the object. So the object must start out moving to the South at 8.9 meters/second.
Its acceleration is +2 m/s^2, which is in the opposite direction to its initial velocity. This means that the velocity of the object changes by +2 m/s every second. After 1 second the velocity of the object will therefore be -8.9 m/s + 2 m/s = -6.9 m/s. After another second the velocity will be -6.9 m/s + 2 m/s = -4.9 m/s. After another second the velocity will be -2.9 m/s, after another -.9 m/s, and after another -.9 m/s + 2 m/s = +1.1 m/s. The speed of the object must therefore decrease, starting at 8.9 m/s (remember speed is always positive because speed doesn't have direction) and decreasing to 6.9 m/s, then 4.9 m/s, etc. until it reaches 0 for an instant, and then starts increasing again.
Since velocities after that instant become positive, the object will therefore start moving to the North immediately after coming to a stop, picking up speed at 2 m/s every second. This will continue until the object has attained a velocity of +20 meters/second and has displaced +80 meters from its initial position.{}{}It is important to understand that it is possible for velocity to be in one direction and acceleration in the other. In this case the initial velocity is negative while the acceleration is positive. If this continues long enough the velocity will reach zero, then will become positive.
STUDENT QUESTION
I understood the negative velocity but was unsure how to explain the rest. I am still rather confused by the last paragraph, expecially where it says that it is possible for velocity to be in one direction and acceleration in the other.
INSTRUCTOR RESPONSE
If you speed up the acceleration is in the direction of motion.
If you slow down the acceleration is opposite the direction of motion.
To speed up a wagon you can get behind it and push in the direction of its motion, giving it an acceleration in its direction of motion.
To slow it down you can get in front of it and push it against its direction of motion (not advisable if it's a big wagon; think of stopping a child in a small wagon), giving it an acceleration in the direction opposite its motion.
STUDENT COMMENTS
Made relative sense, but still unsure in doubt to my answer that the object was increasing after it was moving in the
right direction. Being negative, would it have started towards the south at that acceleration, and moving north would it have
diminished its negativity? It seems this naturally.
INSTRUCTOR RESPONSE
Accelerating to the north (not 'moving' to the north; it does end up moving to the north, but the object starts out moving to the south), the speed in the southward direction would have diminished, as you say.
Eventually it comes to rest, just for an instant, somewhere south of its starting point. Then the northward acceleration will give it an increasing northward velocity.
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Self-critique (if necessary):OK
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Self-critique rating:OK
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