Question

Two cars $A$ and $B$ move such that car $A$ is moving with a uniform velocity of $15\,m\,{\sec ^{ - 1}}$ overtakes car $B$ starting from rest with an acceleration of $3\,m\,{\sec ^{ - 2}}$ . After how much time will they meet again?
A. $5\sec $
B. $15\sec $
C. $\sqrt 3 \sec $
D. $10\sec $

Answer 1

Hint:In order to solve this question, we will use the concept of Newton’s second equation of motion which tells us the distance covered by a body in given time and will compare for both cars $A$ and $B$ to find at what time their distances are equal.

Complete step by step answer:
According to Newton’s second equation of motion, the distance covered by a body in a given time is calculated as $S = ut + \dfrac{1}{2}a{t^2}$ .
Now, let us assume that after time $'t'$ both cars A and B met.
Distance covered by car A in this time is ${S_A} = speed \times t$
speed of car $A$ is given by $15\,m\,{\sec ^{ - 1}}$ so,
${S_A} = 15 \times t \to (i)$
Now, acceleration of car $B$ is given as $3\,m{\sec ^{ - 2}}$ and its velocity is $0$ because it’s starting from rest,
Distance covered by car $B$ in this time is
${S_B} = 0 + \dfrac{1}{2} \times 3 \times {t^2}$
$\Rightarrow {S_B} = 1.5{t^2} \to (ii)$
As, both cars after $t$ time, they will met which means they have covered equal distance
${S_A} = {S_B}$
$\Rightarrow 15t = 1.5{t^2}$
$\therefore t = 10\sec $
So, both cars $A$ and $B$ will meet after a time of $t = 10\sec $ .

Hence, the correct option is D.

Note:Remember, car $A$ was moving with a uniform velocity which means the acceleration of car $A$ was zero and car $B$ starts from rest so its initial velocity $u$ became zero. And From this we can also conclude that car $B$ is travelling faster than $A$ that’s why they met which means an accelerated body travels faster than a body travelling with uniform velocity.