Dragsters can actually reach a top speed ofin onlyconsiderably 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.

• Intial velocity U = 0.
• Final velocity V = 145.
• The taken to reach final velocity T = 4.45 s.

a)

Now to calculate the acceleration, we have the equation of acceleration

$\mathbf{a}\mathbf{=}\frac{\mathbf{V}\mathbf{-}\mathbf{U}}{\mathbf{T}}$

Here V is the final velocity, and U is the initial velocity.

Substituting values in the above expression, we get,

$$\begin{gathered} a=\frac{145-0}{4.45} \\ a=32.6{\text{m/s}}^{\text{2}} \end{gathered}$$

Hence the average acceleration of the dragster is 32.6 m/s².

b)

Data given is:

Distance d = 402.

The dragster's final velocity can be calculated from the equation of motion as:

$V^2-U^2=2ad$

Here V is the final velocity, U is the initial velocity, a is the acceleration, and d is the traveled distance.

Substituting values in the above expression, we get,

$$\begin{gathered} V^2-0^2=2\times 32.6\times 402 \\ V=162\text{m/s} \end{gathered}$$

Hence the final velocity is.

c)

For a dragster, the constant acceleration assumption is invalid.

Various gears on a dragster have different acceleration rates.

The acceleration is strongest in the first few gears and gradually decreases as the gears are increased.

As a result, it would accelerate at a slower pace than in the previous few meters, with a final velocity of less than 162 m/s.