qa10_03Mth

#$&*

course Mth 277

qa 10_03If the velocity function for a projectile is `v(t) = 10 `i + (20 - 9.8 t) `j, then:

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Question: `q001. What is its position function `R(t), and what is its acceleration function `a(t)?

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Your solution:

Int(v(t))=R(t)

R(t)=(10t+c)i+(20t-4.9t^2+c)j

der(v(t))=a(t)

a(t)=-9.8j

confidence rating #$&*:

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Given Solution:

Velocity is the derivative of position, so you need an antiderivative.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question:

`q002. What is its position function if its t = 0 position is `R(0) = 0 `i + 10 `j?

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Your solution:

`R(0) = 0 `i + 10 `j

R(t)=(10t+c)i+(20t-4.9t^2+c)j

R(0)=c1i+c2j

c1=0

c2=10

R(t)=(10t)i+(20t-4.9t^2+10)j

confidence rating #$&*:

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Given Solution:

Your antiderivatives contain integration constants. From the given conditions you can evaluate those constants.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question:

`q003. At what instant is the `j component of the position function equal to 20?

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Your solution:

R(t)=(10t)i+(20t-4.9t^2+10)j

j comp=20

20t-4.9t^2+10=20

-4.9t^2+20t-10=0

Quadr Form Used

J comp is equal to 20 when t approximately equals .5834 and 4.532 units.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question:

`q004. At what instant is the `i component of the position equal to 20, and at that instant what is the `j component of its position?

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Your solution:

R(t)=(10t)i+(20t-4.9t^2+10)j

10t=20

t=2 units

R(2)=20i+30.4j

jcomp=30.4

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question:

`q005. At what instant is the `j component of its position maximized?

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Your solution:

R(t)=(10t)i+(20t-4.9t^2+10)j

jcomp=20t-4.9t^2+10

deriv or v(t)=-9.8t+20=0

t is approx 2.041 units when jcomp os at its max which is approx 30.408

confidence rating #$&*:

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Given Solution:

A function is maximized or minimized at a critical point. A first- or second-derivative test can check whether a critical point gives us a max or a min, or perhaps an inflection point.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question:

`q006. At what instant is the `j component of its position zero, and at that instant what is the `i component of its position?

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Your solution:

R(t)=(10t)i+(20t-4.9t^2+10)j

Once again we need to use quad.

-4.9^2+20t+10=0

one of the times approx equaled -.45 which does not make sense since the time cant be negative so we will discard that answer

the other t was approx equal to 4.53 units which is reasonable and it checks out after plugging it into the original j comp for position.

R(4.532)=45.32i+0j

i comp=45.32

confidence rating #$&*:

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Given Solution:

The quadratic formula might be useful.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question:

`q007. At what instant is the angle between `R(t) and the `i vector equal to 70 degrees? Does this occur at only one instant?

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Your solution:

R(t) dot 'i=

R(t)=(10t)i+(20t-4.9t^2+10)j

'i=(10t)i

'R(t) dot 'i=100t^2

cos(70)=100t^2/(sqrt((10t)^2+(20t)^2+(-4.9t^2)^2+100)sqrt((10t)^2))

cos(70)=100t^2/((49t^3+223.6t^2+100t))

.342=100t/(49t^2+223.6t+100)

16.758t^2+76.47t+34.2=100t

16.758t^2-25.53t+34.2=0

Quadr Form

I must of made an error because I am getting a neg under the radical. I am thinking the a term should maybe be negative, but when I took the magnitude the a term would become positive.

@&

The `R vector is initially directed at an angle of 45 degrees relative to the `i vector. The initial velocity is directed at an angle of less than 70 degrees. So as long as the object is above the x axis the angle can't exceed 70 deg.

Eventually the object will fall below the x axis and the angle of the `R vector with the `i vector will begin increasing. The limiting angle between the vectors will be 90 degrees, and the angle will approach 90 degrees asymptotically.

Regarding your equation, note that (a + b + c)^2 is not equal to a^2 + b^2 + c^2 unless at least two of the values are zero.

*@

confidence rating #$&*:

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Given Solution:

Use the dot product to get an expression for the angle.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question:

`q008. Give a set of parametric equations x = x(t) and y = y(t) that describe the position of the projectile. Eliminate the variable t, and solve for y in terms of x. What kind of equation do you get? Describe its graph.

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Your solution:

R(t)=(10t)i+(20t-4.9t^2+10)j

t=x/10

Now I will plug into y

y=-.49x^2+2x+10

As seen above, we get a quadratice equation and its shape is a parabola which opens downward.

confidence rating #$&*:

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Given Solution:

The position is x(t) `i + y(t) `j. You figured out the position function in the second problem.

You eliminate the variable by solving for either x or y in terms of t, then substituting in the equation for y or x (depending on whether you solve the x or the y equation for t).

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Self-critique (if necessary):OK

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Self-critique rating:OK"

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&#This looks good. See my notes. Let me know if you have any questions. &#