---

Dragsters can actually reach a top speed of 145 m/s in only 4.45 s—considerably less time than given in Example 2.10 and Example 2.11.

(a) Calculate the average acceleration for such a dragster.

(b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402 m (a quarter mile) without using any information on time.

(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:

We are given the following: $v_0=0\ \text{m/s}$; $v_f=145 \ \text{m/s}$; and $\Delta t=4.45 \ \text{sec}$.

Part A

To compute for the average acceleration $a$, we are going to use the formula

Substituting the given values, we have

Part B

We are given the following: $a=32.6 \ \text{m/s}^2$; $v_0=0 \ \text{m/s}$; and $\Delta x=402 \ \text{m}$.

Since we do not have any information on time, we are going to use the formula

To compute for the final velocity, we have

Substituting the given values:

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  $162\:\text{m/s}$.

---