# College Physics by Openstax, Problem 2.37

#### (c) Why is the final velocity greater than that used to find the average acceleration? Hint: Consider whether the assumption of constant acceleration is valid for a dragster. If not, discuss whether the acceleration would be greater at the beginning or end of the run and what effect that would have on the final velocity.

SOLUTION:

Part a

The average acceleration is

$a=\frac{v-v_0}{t}$

$a=\frac{145\:m/s-0\:m/s}{4.45\:s}$

$a=32.6\:m/s^2$

Part b

The final velocity is

$v^2=v_0^2+2ax$

$v=\sqrt{v_0^2+2ax}$

$v=\sqrt{\left(0\:m/s\right)^2+2\left(32.6\:m/s^2\right)\left(402\:m\right)}$

$v=162\:m/s$

Part c

The final velocity is greater than that used to find the average acceleration because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears, and would have a greater acceleration in first gear than second gear than third gear, etc. The acceleration would be greatest at the beginning, so it would not be accelerating at $32.6\:m/s^2$ during the last few meters, but substantially less, and the final velocity would be less than  $162\:m/s$.

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