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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 -

## 5 Answers

7 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.

7 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.

7 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

9 months 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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