College Physics by Openstax, Problem 2.37

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:

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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