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 -

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.

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.

  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

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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