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

a=\frac{\Delta v}{\Delta t}=\frac{v_f-v_0}{\Delta t}

Substituting the given values, we have

\begin{align*}
a & =\frac{v_f-v_0}{\Delta t} \\
a & =\frac{145\:\text{m/s}-0\:\text{m/s}}{4.45\:\text{s}} \\
a & =32.6\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

To compute for the final velocity, we have

v_f=\sqrt{\left(v_0\right)^2+2a\Delta \:x}

Substituting the given values:

\begin{align*}
v_f & =\sqrt{\left(v_0\right)^2+2a\Delta \:x} \\
v_f & =\sqrt{\left(0\:\text{m/s}\right)^2+2\left(32.6\:\text{m/s}^2\right)\left(402\:\text{m}\right)} \\
v_f & =162\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}

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}.