Question

[AS7] A man is 48 m behind a bus which is at rest. The bus starts accelerating
the rate of 1 m/s. At the same time the man starts running with uniform velocity of
m/s. What is the minimum time in which the man catches the bus? [Refer to TB page
3.6 Higher Order Thinking Skills
Q1.
2
33 03]
A.
B В
18​

Answer 1

Answer:

Explanation:

Given,

Acceleration of bus, a=1ms  

−2

 

Assume man catches the train in timet.

Displacement by man in time t s  

1

​  

=vt=10t

Displacement by bus in time t, s  

2

​  

=ut+  

2

1

​  

at  

2

=  

2

t  

2

 

​  

 

Separation between man and bus is 48m

s  

1

​  

=s  

2

​  

+48

10t=  

2

t  

2

 

​  

+48

t  

2

−20t+96=0

t=  

2

20±  

20  

2

−4×96

​  

 

​  

 

t=8and12

In first attempt he caches the bus

Hence, minimum time he catches the bus is 8sec