Question

A purple car is moving three times as fast as a yellow car. Each car slows down to a stop with the same constant acceleration. How much more distance is required for the purple car to stop ?

• A. twice as much distance
• B. three times as much distance
• C. five times as much distance
• D. seven times as much distance
• E. nine times as much distance

Answer 1

The correct option is E nine times as much distance

Let the initial velocity of purple car be $3u$ , so the initial velocity of yellow car will be $u$. Final velocities ( $v_p,v_y$ )of both cars are zero , acceleration $a$ is same for both cars .

Using, the third equation of motion $v^2=u^2-2as$

For purple car $0=(3u{)}^2-2a{s}_p$

where $s_p=$ distance required for purple car to stop ,

$a=$ acceleration (deceleration)

Hence, $s_p=9u^2/2a$ ,

For yellow car $0=u^2-2a{s}_y$

Hence, $s_y=u^2/2a$

where $s_y=$ distance required for yellow car to stop ,

now by dividing $s_p$ from $s_y$

we get $s_p/s_y=9$

or $s_p=9s_y$

Hence, the purple car requires nine times as much distance as the yellow car to stop.