Q. A bus starts from rest with an acceleration of 1m/s2. A man who who is 48m behind the bus starts with a uniform velocity of 10m/s. Calcualate the minimum time after which the man will catch the bus.
Answers

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.
4 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.
4 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
4 years ago

This is your answer of bus starts from rest with an acceleration of 1m/s2. A man who who is 48m behind the bus starts with a uniform velocity of 10m/s. Calcualate the minimum time after which the man will catch the bus.
8 Seconds
Given:
Acceleration of the bus = 1m/s^2
Distance of the man = 48 meters
For solving this question we have to frame an quadratic equation:
The equation must be framed in the following way:
= Distance Covered by the Bus = Distance Covered by the man - 48
Taking these values in terms of physics:
1/2 a t^2 = vt - 48
Substituting all the values known to us in this formula:
1/2(1)t^2 = 10t - 48
Multiplying the Equation by 2 to simplify:
t^2 = 20t -96
Taking the LHS to the RHS:
t^2 - 20t + 96
We get a quadratic equation now solving this equation by splitting the middle term:
t^2 - 12t - 8t + 96
Taking t and - 8 common we get:
t (t - 12) - 8(t -12)
(t - 12) and (t -8)
t = 12 sec and t = 8 sec
Therefore the time taken can either be 12 seconds or 8 seconds but here in this question they asked the minimum time after which the man will catch the bus so this would be the less time of the two so the answer is 8 seconds.