Question

A bus begins to move with a=1 m/s^2 a man who is 48 metre behind the bs starts running at 10 m/s to catch the bus then men will be able to catch the bus after how many seconds?

Answer 1

We need to use the  concept of the Relative motion here. Now,

Acceleration of Bus = 1 m/s²

Acceleration of man = 0

Relative acceleration of man w.r.t bus = 0 - 1 = -1 m/s².

Initial velocity of man = 10

Initial velocity of the Bus = 0 m/s.

Relative initial velocity of man w.r.t bus = 10 - 0 = 10 m/s.

Relative distance = 48 m.

Using the Equation of the motion,

S = ut + 1/2 × at²

⇒ 48 = 10t + 1/2× -1t²

⇒ 96 = 20t - t²

∴t² - 20t + 96 = 0

On Solving this quadratic equation, we will get,

t = 12 or t = 8 seconds.

Hope it helps.