course Phy 201
I had some trouble with this assignment but feel i will understand by the end of this section.
006. Using equations with uniformly accelerated motion.Question: **** `q001. Note that there are 9 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=vo+a*dt
vf-vo=a*dt
(vf-vo)/dt=a
(30m/s-10m/s)/15s=a
(20m/s)/15s=a
1.33m/s^2=a
Confidence Assessment: 3
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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):i understood this problem
Self-critique rating:
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:
you know the inital and final velocity so you would take the difference of those two velocities and divide them by the number of seconds it takes. this gives you the rate of change in the velocity or the acceleration.
Confidence Assessment:
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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.
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Self-critique (if necessary):none
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+vo/2)*dt
(2/dt)*ds=vf+vo
-vf*ds*(2/dt)=vf+vo
ds*(2/dt)-vf=vo
(80m*2/10s)-6m/s=vo
(160m/10s)-6m/s=vo
16m/s-6m/s=vo
10m/s=vo
Confidence Assessment: 2
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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):i understand this problem
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:
Because you are given ds and dt you can find vAve.by dividing ds by dt. Once given vAve, must find the average between the two points to get the Vo.
Confidence Assessment: 2
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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):i understand this problem.
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:
dt=10s
vf=?
a=/2m/s^2
vo=?
ds=80m
dt=10s
ds=vo*dt+.5a*dt^2
ds/(dt+.5a*dt^2)=vo
The above step is where you went wrong. `dt is multiplied by v0, but .5 a `dt^2 is not, so the latter expression should not have been part of your denominator.
As shown in the given solution
`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.
80m/ 10s+.5(-2m/s^2)*10s=vo
80m/10s+(-1m/s^2)*10s=vo
80m/10s-10s^2=vo
80m/s=vo
vf=vo+a*dt
=80m/s+-2m/s^2
=800m-2m/s^2
vf= 798m/s
Confidence Assessment:
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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):i understand where i went wrong in my equation. I am having some trouble understanding where to break off the orignal equation. If you could go into further detail that would be great.Self-critique rating:
rating: 1
You had one erroneous step in your algebra. You didn't completely work out the units right (they would have been inconsistent because of your algebra error), but overall your approach was very good.
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: i dont really understand this problem.
Confidence Assessment: 0
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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):i dont understand how you are using direct reasoning if the are equations.
This solution verifies the solution obtained by equations, using direct reasoning from the definitions of velocity and acceleration. It takes the results of the preceding solution and demonstrates in a direct manner that they are consistent.
&#This looks good. See my notes. Let me know if you have any questions. &#