Question

A car starts from rest and accelerates uniformly at 3.0 m/s2. A second car starts from rest 6.0 seconds later at the same point and accelerates uniformly at 5.0 m/s2. How long does it take the second car to overtake the first car?

Answer 1

I'm using 1/2at^2 = 1/2a(t-6)^2 but getting confused somewhere in the math.

Answer 2

I don't see an issue with what you're doing...you should end up with a value of 26.6 secs. Maybe if you show your calcs someone can chime in a figure out what's going on (logging off soon).

Answer 3

3/2t^2 = 5/2(t^2-12t+36)
3/2t^2 = 5/2t^2-30t+90
t^2 = 5/3t^2-20t+60
Still working on the rest!

Answer 4

you can think an other way
X1 = 1/2a1*t^2+vo*t + xo , vo = 0 , xo = 1/2a1*t1*2=24
X1 = 1/2a1*t^2 + 24
X2 = 1/2a2 * t^2
second car over take first car unless X1 = X2
1/2a1*t^2 + 24 = 1/2a2*t2