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Q- A bus starts from rest with an acceleration of 1 m/sec2. A man who is 48 m behind the bus starts with a uniform velocity of 10 m/sec, then the minimum time after which the man will catch the bus is -

Q- A bus starts from rest with an acceleration of 


1 m/sec2. A man who is 48 m behind the bus 


starts with a uniform velocity of 10 m/sec, then 


the minimum time after which the man will 


catch the bus is -

Grade:

5 Answers

Ritwik Kala
26 Points
10 years ago

ANS - 8 seconds

In BUS FRAME-

velocity of man = 10 m/s

s = 48 m

Acceleration of man with respect to bus = Acceleration of man - Acceleration of bus = 0 - (1) = -1 m/s

Applying second equation of motion,

s = ut + 1/2 at2

48 = 10t - 1/2*t2

solving we get, t = 12 sec or t  = 8 sec

therefore the minimum time is 8 seconds.

Mujahid Ahmed
42 Points
10 years ago

Let S be the distance travelled by Bus when they meet.

Now,

 

S = S + 48

1/2 x 1 x t2 = 10 x t + 48.

 

Solve for t.

 

Approve if i helped.

kshitij aggarwal
18 Points
10 years ago

  Distance travelled by bus = Distance travelled by man - 48

 

so 1/2 X 1 Xt2 = 10t - 48

t2 - 20t + 96 = 0 

 

t = 8sec, 12 sec

 

earlier time is 8 seconds

Pranshu
12 Points
6 years ago
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Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

Let man will catch the bus after time t.Since the distance from man position is same for both at time t
⇒ Distance covered by bus =Distance covered by man -48
⇒ 0.5​at^2 = vt − 48
⇒ 0.5​(1)t^2 = 10t − 48
⇒ t^2 − 20t + 96 = 0
⇒ (t−12)(t−8) = 0
⇒ t=8s (minimum value) or t=12s
So,at t=8s,the man will catch the bus

Thanks and Regards

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