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November 28th, 2015, 01:07 AM   #1
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how to find the distance given the velocity?

Starting at rest, the velocity of an object is given by v = 3t^2. Find the distance traveled after 5 seconds.

My solution:

d =vt

but v = 3t^2

so: d = (3t^2)(t)

@t=5

d = [3(5)^2](5)
d = [3(25)](5)
d = 75(5)
d = 375 units

My answer is obviously correct,

but the answer in the book is d=125.

Please tell me whether the book is wrong or I am wrong,
If I am wrong, please explain why I am wrong... thank you.

Last edited by skipjack; November 28th, 2015 at 01:35 AM.
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November 28th, 2015, 01:47 AM   #2
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You are wrong. The equation d = vt applies if v is the mean velocity. In particular, this equation is correct if v is a constant.

You are given that v = 3t². You need to use the fact that v is the derivative of d with respect to time.

As d = 0 when t = 0, d = t³, and so d = 125 when t = 5.
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