Mechanics : Equations of motion.

1. Apr 28, 2010

Maatttt0

1. The problem statement, all variables and given/known data

A motorcyclist starts from rest at a point O and travels in a straight line. His velocity after t seconds is vms^-2, for 0 =< t =< T, where v = 7.2t - 0.45t^2. The motorcyclist's accelaration is zero when t = T.

Find the value of T.

2. Relevant equations

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3. The attempt at a solution

S = X
U = 0
V = 7.2t - 0.45t^2
a = 0
t = t

v = u + at
7.2t^2 - 0.45t = 0
0.45t(t - 16) = 0
Therefore t = 0 or 16

But the answer is 8.. I believe I'm getting the accleration part incorrect but I just cannot spot it.

Last edited: Apr 29, 2010
2. Apr 28, 2010

hotvette

The expression for v you used in step 3 doesn't match the problem statement. You swapped terms.

3. Apr 29, 2010

Maatttt0

Ahh apologies - I mistyped it. The second line is supposed to have the ^2 on the 0.45t :(

4. Apr 29, 2010

HallsofIvy

Staff Emeritus
This is only true for a constant acceleration- and a constant acceleration gives a linear velocity, not quadratic as you have here.

All you have done here is solve for V= 0, not acceleration.

The acceleration is the derivative of the velocity function. Take the derivative of $7.2t^2- 0.45t$ and set that equal to 0.

5. May 2, 2010

Maatttt0

Thank you for the reply. I remember my teacher had mention not to attempt that question as he had not covered it in class yet. I read up about it and you're help has reinforced my understanding.

Thanks again. :)