A bus starts from rest with an acceleration of 1m/s^2 . A man behind the bus runs at a speed of 10m/s and the bus is initially 48m away from the man how much time he will take to reach the bus

Question

kanwal32 · Answer

@mathmale hlp

kanwal32 · Answer

@ganeshie8 nd @hartnn hlp doing this type of question first time

kanwal32 · Answer

with constant veloctiy

mathmale · Answer

Think of some starting point, e. g., x=0 feet.  The bus starts from this point and moves in the positive x-direction.  The man starts not at x=0 but at x=48 feet.  If the man manages to catch up with the bus, then the man and the bus are the same distance from x=0 feet, right?  

So:  

Write an expression for the distance traveled by the bus as a function of time.
Then write one for the distance traveled by the man as a function of time.

Equate these two expressions and solve the resulting equation for time, t.

kanwal32 · Answer

its relative motion

kanwal32 · Answer

i'll solve by myself

mathmale · Answer

Hint:$$s=s _{0}+v _{0}t+\frac{ 1 }{ 2 }at^2$$

mathmale · Answer

What is 's0' for the bus?  for the man?  What is 'a' for the bus?  for the man?

kanwal32 · Answer

a is zero for the man cause constant velocity