Question

Consider two toy cars. Car A starts from rest and speeds up with constant acceleration for a time delta t until it reached a speed of v and then continues to travel at this speed. At the moment car A reaches its maximum speed, car B, starting at rest from the same point that car A started from, speeds up with constant acceleration.

Determine how much time elapses between the time car B starts moving and the time car B passes car A. To do this you will need to solve a quadratic equation?

Answer 1

At time t > ∆t,
position of A: v(t-∆t) + 1/2 (v/∆t)t^2
position of B: 1/2 a(t-∆t)^2

So, now you need to find when the positions are the same:

v(t-∆t) + 1/2 (v/∆t)t^2 = 1/2 a(t-∆t)^2
2vt - 2v∆t + (v/∆t)t^2 = at^2 - 2a∆t t + a(∆t)^2

Now just collect terms and plug in your values for a,v,∆t

Time t-∆t elapses after car B starts.

Answer 2

car A:
v=at1 or t1=v/a
total distance traveled
d=1/2 a*t1^2+v(t2) = 1/2 (v^2/a) + v*t2
car B:
distance=1/2 ab*t2^2

but the distance they travel are the same,

1/2 ab*t2^2 =1/2 (v^2/a)+v*t2

Now the problem we have is what is accelearations a, and ab? You cant solve for three unknowns (a, ab, t2) with one equation. So I am assumeing a=ab (ie, the cars accelerations is identical)

then
a*t2^2=(v^2/a)+v*t2 this makes it easy..
a*t2^2-v*t2 -v^2/a=0

and you can use the quadratic equation to solve for t2 in terms of a.

Answer 3

At bobpursley, why do you have 1/2(v^2/a)?

Answer 4

distance is 1/2 a t^2, in this case t=V/a) recheck it, I do periodically make errors when typing math.

Answer 5

Thanks a lot man, I really appreciate it