Question

A professional race-car driver buys a car that can accelerate at 5.2 m/s2 . The racer decides to race against another driver in a soupedup stock car. Both start from rest, but the stock-car driver leaves 1.3 s before the driver of the sports car. The stock car moves with a constant acceleration of +3.2 m/s2 . Find the time it takes the sports-car driver to overtake the stock-car driver. Answer in units of s.

Answer 1

Answer:

6.02 s

Explanation:

We can write the position of the stock car as:

$x_1 = \frac{1}{2}a_1 t^2$

where

$a_1 = 3.2 m/s^2$ is the acceleration of the stock car.

The sport car instead starts its motion only 1.3 s afterwards, therefore its position at time t can be written as

$x_2 = \frac{1}{2}a_2 (t-1.3)^2$

where

$a_2 = 5.2 m/s^2$ is the acceleration of the sport car

(we can verify indeed that when t = 1.3 s, $x_2 = 0$).

The sport car reaches the stock car when the two positions are equal:

$x_1 = x_2 \\\frac{1}{2}a_1 t^2 = \frac{1}{2}a_2 (t-1.3)^2$

Rewriting the equation,

$a_1 t^2 = a_2 (t-1.3)^2\\3.2t^2 = 5.2(t-1.3)^2 = 5.2t^2-13.5t+8.8\\2t^2-13.5t+8.8 =0$

This is a second-order equation with two solutions:

t = 0.73 s

t = 6.02 s

We discard the first solution since we are only interested in the times > 1.3 s, therefore the sport car overcomes the stock car after

6.02 seconds.