Question

A car and bike start racing in a straight line. The distance of the finish line from the starting line is 100m. The minimum acceleration of car to win, if it accelerates uniformly starting from rest and the bike moves with a constant velocity of 10 m/s, is

A:0.5m/{s^2} \\\ B:2m/{s^2} \\\ C:1m/{s^2} \\\ D:3m/{s^2} \\\ \end{gathered} $$

Answer 1

Acceleration can be defined as the rate of change of velocity of any object with respect to time. And uniform acceleration means that acceleration of an object remains same or constant irrespective of the time. An example of a uniformly accelerated motion is a ball rolling down a slope.

Step by step solution: For a uniformly accelerated motion, following equations are taken into consideration:

$\begin{gathered} v = u + at{\text{ (1)}} \\\ s = \dfrac{1}{2}(u + v)t{\text{ (2)}} \\\ s = ut + \dfrac{1}{2}a{t^2}{\text{ (3)}} \\\ s = vt - \dfrac{1}{2}a{t^2}{\text{ (4)}} \\\ {v^2} = {u^2} + 2as{\text{ (5)}} \\\ \end{gathered}$

The aforementioned equations cannot be applied in cases when acceleration is not constant.

In the question, we are given a uniformly accelerated motion of a bike and a car which are moving in a straight line. We are provided with the following information:

In case of bike:

Velocity, v=10 m/s
Distance, D = 100 m

Time, $t = \dfrac{D}{v} = \dfrac{{100}}{{10}} = 10s$

In case of car in order to win the race:
Time, t = 10 s
Initial velocity, u = 0
S = 100
Using the equation (3):
$100 = 0 \times t + \dfrac{1}{2}a{\left( {10} \right)^2}$
$a = 2m/{s^2}$

Thus, the minimum acceleration of a car to win, if it accelerates uniformly starting from rest and the bike moves with a constant velocity of 10 m/s, is 2 m/s 2 .

Hence, the correct answer is Option B.

Note: If an object's speed or velocity is increasing at a constant rate then it indicates that it possesses uniform acceleration which means rate of acceleration is constant. But if a car speeds up and then slows down and then again speeds up, it indicates that it doesn't has a uniform acceleration.