A man runs at a speed of `4m//s` to overtake a standing bus. When he is `6m` beind the door at `t=0`, the bus mover forward and continuous with a constant acceleration of `1.2m//s^(2)`. The man reaches the door in time `t`. Then

Options

A. 5.2 s
B. 4.3s
C. 2.3s
D. the man shall never gain the door

Answer

C

Explanation

At t=0 let the man's position be the origin.
`therefore x_(m0)=0`
The bus door in then at `x_(b0)=6.0m`.
The equation of motion for the man is
`x_(m)=x_(m0)+v_(m0)t+(1)/(2)a_(m)t^(2)`
Here, `x_(m0)=0,v_(m0)=4.0m//s, a_(m)=`
`therefore x_(m)=4t` ...........(i)
The equation of motion for the bus is
`x_(b)=x_(b0)+V_(b0)t+(1)/(2)a_(b)t^(2)`
Here, `x_(b0)=6.0m, v_(b0)=0,a_(b)=1.2 m//s2`
`therefore x_(b)=6+(1)/(2)(1.2t^(2),x_(b)=6+0.6t^(2)` ...........(ii)
When the man catches the bus,
`x_(m)=x_(b)`
`therefore 4t=6+0.6t^(2)` (Using (i) and (ii))
or `0.6t^(2)-4t+6=0`
This can be reexpressed as
`3t^(2)-20t+30=0`
Solving this quadratic equation by quadratic formula, we get
`t=(20+-sqrt(400-360))/(6)=(10+-sqrt(10))/(3)2.3s, 4.4 s`.