Question

A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration $2m/{s^2}$and the car has acceleration $4m/{s^2}$. The car will catch up with the bus after a time of:
A. $10\sqrt 2 s$.
B. $\sqrt {110} s$
C. $\sqrt {120} s$
D. $15s$

Answer 1

HintIn order to solve the problem use the concept of relative motion that when two objects are moving simultaneously then one motion of one body is observed with respect to another body. First we will find the relative acceleration of the car with respect to the bus, since both are moving in the same direction therefore we will take the difference of their acceleration. Afterwards we will use the second equation of Newton to determine the time.

 Complete step-by-step solution:
According to initial data we have
Initial displacement between the bus & car, $s$= 200m
The acceleration of bus, ${a_b} = 2m/{s^2}$
The acceleration of car, ${a_c} = 4m/{s^2}$
We know that,
When two bodies A & B are moving in same direction then
Relative velocity, Velocity of A with respect to B i.e. motion of A is observed from B
${v_{AB}} = {v_A} - {v_B}$
Relative acceleration, Acceleration of A with respect to B i.e. motion of A is observed from B
${a_{AB}} = {a_A} - {a_B}$
When two bodies A & B are moving in opposite direction then
Relative velocity, Velocity of A with respect to B i.e. motion of A is observed from B
${v_{AB}} = {v_A} + {v_B}$
Relative acceleration, Acceleration of A with respect to B i.e. motion of A is observed from B
${a_{AB}} = {a_A} + {a_B}$
Therefore, relative acceleration of car with respect to bus, ${a_{cb}}$ = ${a_c} - {a_b}$ = $4 - 2$ = $2$
Let the time taken by the car to catch the bus = $t$
Let the initial velocity of car = ${u_c}$ =0
Let the initial velocity of bus = ${u_b}$ =0
Relative velocity of car with respect to bus = ${u_{cb}}$ = ${u_c} - {u_b}$ = 0
Using second equation of motion
$s = {u_{cb}}t + \dfrac{1}{2}{a_{cb}}{t^2}$
On putting the values we have
\[\Rightarrow 200 = 0 \times t + \dfrac{1}{2}(4 - 2){t^2}\]
\[\Rightarrow 200 = \dfrac{1}{2} \times 2{t^2}\]
On simplifying we get
\[\Rightarrow {t^2} = 200\]
\[\Rightarrow t = \sqrt {200} \]
Finally we get
\[\Rightarrow t = 10\sqrt 2 s\]
Hence the car will catch up with the bus after a time, \[t = 10\sqrt 2 s\]
Therefore the correct option is A.

Note:This problem can also be solved by another method.
Method : Assume the distance travelled by bus = ${s_b}$ = $x$
As the car will catch the bus standing 200m behind then assume distance travelled by car = ${s_c}$= $200 + x$
Acceleration of the bus, ${a_b} = 2m/{s^2}$
Acceleration of the car, ${a_c} = 4m/{s^2}$
Both car & bus start from rest ${u_c}$& ${u_b} = 0$
Let the time when the car will catch the bus = $t$
Using second equation of motion for bus
${s_b} = {u_b}t + \dfrac{1}{2}{a_b}{t^2}$
Using second equation of motion for car
${s_c} = {u_c}t + \dfrac{1}{2}{a_c}{t^2}$
Subtracting ${s_c}$ & ${s_b}$
\[{s_c} - {s_b} = {u_c}t + \dfrac{1}{2}{a_c}{t^2} - ({u_b}t + \dfrac{1}{2}{a_b}{t^2})\]
\[\Rightarrow 200 + x - x = 0 + \dfrac{1}{2}{a_c}{t^2} - 0 - \dfrac{1}{2}{a_b}{t^2}\]
\[\Rightarrow 200 = \dfrac{1}{2}(4 - 2){t^2}\]
\[\Rightarrow 200 = \dfrac{1}{2}(4 - 2){t^2}\]
\[\Rightarrow 200 = \dfrac{1}{2} \times 2{t^2}\]
\[\Rightarrow {t^2} = 200\]
\[\Rightarrow t = \sqrt {200} \]
\[\Rightarrow t = 10\sqrt 2 s\]