Question

Find the initial velocity and constant acceleration of the green car?

Answer 1

Let initial velocity of green car be`=-u_ghatx`
Also let its acclerartion be `=-ahatx`
Given that at `t=0, vecx_g-vecx_r=224m`
Origin is at `x_r`

1. Red car having constant velocity`=28.0xx1000/3600=7.bar7ms^-1`
   Let the two cars meet after time `t_1` at a given displacement of `43.4hatx`.
   `=>` Displacement of red car is `=43.4hatx` in time `t_1`
   `:.t_1=43.4/(7.bar7)=5.58s`
2. Red car having constant velocity`=56.0xx1000/3600=15.bar5ms^-1`
   Let the two cars meet after time `t_2` at a given displacement of `76.1hatx`.
   `=>` Displacement of red car is `=76.1hatx` in time `t_2`
   `:.t_2=76.1/(15.bar5)=4.892s`

Displacement of green car in first case `=(43.4-224)hatx` `=-180.6hatx`
Setting up the equation for green car using kinematic expression
`s=ut+1/2at^2` we get for `t_1`
`(-u_g xx5.58+1/2(-a5.58^2))hatx=-180.6hatx`

`=>5.58u_g +15.5682a=180.6` .........(1)

Displacement of green car in second case `=(76.1-224)hatx` `=-147.9hatx`
Setting up the equation for green car in the second case
`(-u_g xx4.892+1/2(-a4.892^2))hatx=-147.9hatx`

`=>4.892u_g+11.966a=147.9` ......(2)

Solve simultaneous equation (1) and (2) to calculate `u_gand a`
Hint: You can divide (1) with `5.58` and (2) with `4.892` and subtract one from other, thus eliminating `v_g` to find out value of `a`. Then insert the obtained value of `a` in either of the two equations to calculate `v_g`.

Report the value of velocity and acceleration of green car as assumed initially.