Question

Initially car A is 21 m ahead of car B. Both start moving at t=0 in same direction along straight line. If A is moving with constant velocity 20
m/s and B starts from rest with an acceleration 2 m/s2 towards A, then the time when car B will catch the car A is
O 21 s
0 22 s
O 10 s
O Can never catch A​

Answer 1

Given:

Initial distance between car A and car B= 21 m

At t=0

Car A is ahead of car B

Initial velocity of car A= 20 m/s

Initial velocity of car B= 0 m/s

(Since, car B starts from rest)

Acceleration of car B= 2 m/s²

To Find:

Time taken by car B to catch car A

Solution:

We know that,

$\pink{\underline{\boxed{\bf{Distance=Speed\times time}}}}$

• According to second equation of motion for constant acceleration,

$\purple{\underline{\boxed{\bf{s=ut+\frac{1}{2}at^{2} }}}}$

where,

u is initial velocity

a is acceleration

s is displacement

t is time taken

━━━━━━━━━━━━━━━━━━━━━━━━━

Suppose car B will catch car A in time 't'

So, let the distance covered by car A in time t be x

So,

$\longrightarrow\rm{Distance=Speed\times time}$

$\longrightarrow\rm{x=20t}------(1)$

Also, the distance covered by car B after time t will be 21+x

So, on applying second equation of motion on car B, we get

$\longrightarrow\rm{s=ut+\frac{1}{2}at^{2}}$

$\longrightarrow\rm{21+x=0(t)+\dfrac{1}{2}(2)t^{2}}$

$\longrightarrow\rm{21+x=t^{2}}$

From (1), we get

$\longrightarrow\rm{21+20t=t^{2}}$

$\longrightarrow\rm{0=t^{2}-20t-21}$

$\longrightarrow\rm{t^{2}-20t-21=0}$

$\longrightarrow\rm{t^{2}-21t+t-21=0}$

$\longrightarrow\rm{t(t-21)+1(t-21)=0}$

$\longrightarrow\rm{(t+1)(t-21)=0}$

$\longrightarrow\rm{t=-1,21}$

Since, time cannot be negative

Therefore,

$\longrightarrow\rm\green{t=21\ s}$

Hence, the time taken by car B to catch car A is 21 s.