# College Physics 2.37 – The acceleration of dragsters

## Solution:

### Part A

The average acceleration is

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

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

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

### 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/s2 during the last few meters, but substantially less, and the final velocity would be less than  $162\:m/s$.