A car starts from a certain point at a velocity of 100 kph. After 20 seconds, another car starting from rest, starts at the same point following...

Hello!

Certainly car B will overtake car B. After some time the speed of car B will become greater than the speed of car A and a quicker object always overtakes a slower one.

Now determine the specific time. The distance travelled by car A is

`d_A=V_A*t,` where `V_A=100 (km)/h=(100,000m)/(3600s) approx 27.8m/s` and time `t` is measured from the start.

The distance travelled by car B is

`d_B=a_B*(t-20)^2/2,` where `a_B=3m/s^2.`

We have to find `t_1gt20` such that `d_A(t_1)=d_B(t_1):`

`27.8*t=3(t-20)^2/2,` or

`3t^2-120t-2*27.8t+1200=0,` or `3t^2-175.6t+1200=0.`

So `t_1=(175.6+sqrt((175.6)^2-4*3*1200))/6 approx` 50.6 (s). The solution with minus before the root is less than 20.

The distance both travel will be `d_A=27.8*50.6 approx` 1407 (m).