Specialist Kinematics Help Please

[Brace]

i have this question and i have been stuck on it for days now please help

Cars A and B are stationary on a straight road, and A is 21 metres in front of B. Car A moves off, away from B, with an acceleration 1ms^-2, which is maintained for 20 seconds, after which the car continues to move at a constant speed.

B sets off 10 seconds after A starts, in pursuit, with a constant acceleration of 2ms^-2 until it draws level it A. Find toe taken by B in pursuit and the distance B travels

i know the answer is 21 seconds and 441 metres but how do i get there?

Cheers Brace

[lzxnl]

OK, so let B's initial position be the origin. A's position is given by 21 + 0.5  t^2 for t<20. At t = 20, A has moved 221 metres, and its speed is 20 m/s, so for t >= 20, A's position = 221 + (t-20) * 20

B's position is simply (t-10)^2 as distance = 1/2*aT^2 = (t-10)^2 due to its 10 second delay

Firstly, try (t-10)^2 = 21 + 0.5 t^2 , i.e. t<=20

We get t^2 - 20t + 100 = 21 + 0.5 t^2
0.5 t^2 -20t + 79 = 0
t = 20 +- sqrt(400-2*79) = 20 +- sqrt(242) which either way is out of the domain [0,20]

So try (t-10)^2 = 221 + 20t - 400 = 20t - 179
t^2 - 40t + 279 = 0
(t-20)^2 +279 = 400
(t-20)^2 = 121
t = 20+-11 = 31 or 9. Use t = 31

Subbing into B's distance of (t-10)^2, (t-10)^2 = 21^2 = 441 m

I'm not sure t = 21 s is correct.