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?
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.
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.
At bobpursley, why do you have 1/2(v^2/a)?
distance is 1/2 a t^2, in this case t=V/a) recheck it, I do periodically make errors when typing math.
Thanks a lot man, I really appreciate it
To solve this problem, we can break it down into four different parts and then combine them to find the total time elapsed between car B starting and passing car A.
1. Time taken by car A to reach its maximum speed:
Using the equation of motion, v = u + at, where v is the final velocity, u is the initial velocity (0 in this case), a is the acceleration, and t is the time taken, we have:
v = 0 + a(t_A)
t_A = v/a
2. Distance traveled by car A to reach its maximum speed:
Using the equation of motion, s = ut + 1/2at^2, where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time taken, we have:
s_A = 0*t_A + 1/2a(t_A)^2
s_A = 1/2a(t_A)^2
3. Time taken by car B to cover the distance traveled by car A:
As car B starts from rest and speeds up with constant acceleration, we can use the equation of motion, v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled, we have:
v_B^2 = 0 + 2a(s_A)
v_B^2 = 2as_A
v_B = sqrt(2as_A)
t_B = v_B/a
4. Time elapsed between car B starting and passing car A:
The total time elapsed between car B starting and passing car A is the sum of the time taken by car A to reach its maximum speed and the time taken by car B to cover the distance traveled by car A:
delta t = t_A + t_B
Now, you mentioned that we need to solve a quadratic equation. This occurs because in step 3, we used the equation v_B^2 = 2as_A, which can be rearranged as a quadratic equation: at^2 - 2s_A = 0. Solving this equation will give us the value of t_B.
Note: The actual values of v, a, and s_A need to be given in order to obtain the numerical solution for delta t.