Question

Speeds of two identical cars are $u$ and \[4u\] at a specific instant. What is the ratio of the respective distance at which the two cars are stopped from that instant?

Answer 1

Hint: motion of an object is associated with some reference point. When we say that some object is in motion, we mean that the distance of the object from the reference or starting point is changing. The moving objects always have some speed and direction known as velocity.

Formula used: \[{v^2} - {u^2} = 2as\]
Here, $a$ is the acceleration, $s$ is the displacement, $v$ is the final velocity, and $u$ is the initial velocity.

Complete step by step answer:
 It is given that the speeds of two cars at some instant are $u$ and \[4u\]. We have to find the distances covered by each of them when they come to rest and take their ratios.
Now, we know that the car which is having greater speed will travel a large distance as compared to the one having less velocity when they are stopped with the same deceleration.
Initial velocities are given and final velocities will be zero for both, i.e., $v = 0$
Let us substitute the values in the following equation of motion.
\[{v^2} - {u^2} = 2as\]
${0^2} - {u^2} = 2( - a)s$
So, we get
${u^2} = 2as$
$s = \dfrac{{{u^2}}}{{2a}}$ ------- (1)
Similarly for another car,
${0^2} - {(4u)^2} = 2( - a)s$
$16{u^2} = 2as'$
$s' = \dfrac{{16{u^2}}}{{2a}}$ --------- (2)
Now let us find the ratio of their distances by dividing equations (1) and (2).
$\dfrac{s}{{s'}} = \dfrac{{{u^2}/2a}}{{16{u^2}/2a}}$
$ \Rightarrow \dfrac{s}{{s'}} = \dfrac{1}{{16}}$
Therefore, the ratio of the distances covered by the two cars is $1:16$.

Note:
When a car/object starts from rest we take its initial velocity to be zero. Similarly, when a car/object comes at rest we take its final velocity to be zero.
Acceleration is a term used when the speed of an object increases and deceleration is the term used when the speed of an object decreases.