Assumptions:
The blue car travels from a dead stop with constant acceleration and reaches a peak velocity of 20 m/s after traveling 100 m. It then proceeds to travel at a constant velocity of 20 m/s after that.
The red car starts from a dead stop 3 seconds later and travels at a constant acceleration of 2.5 m/s2 to a peak velocity 22 m/s. It the proceeds to travel at a constant velocity of 22 m/s after that.
Question: When does the red car catch up to the blue car and how far have the cars traveled when they meet?
Procedure:
Since the distance each car travels can be expressed as a function of time (actually a piecewise function but we will get to that), we want to get x(t) expressions (one for the blue car and one for the red car) and then set them equal to each other. We can then solve for t (time) and then put that back into the expressions to get x (distance).
We will be using these two equations of motion:
Position: x(t) = 1/2 a (t-t0)2 + v0(t-t0) + x0
Velocity: vf2 = v02 + 2a(xf - x0)
Where:
x0 = initial position
xf = final position
v0 = initial velocity
vf = final velocity
t0 = initial time
a = acceleration
Let's try to get x(t) for the blue car.
Blue Car Acceleration Phase
But first we need to realize that we don't know the acceleration nor the duration for the period of time that the blue car is accelerating.
So we will use the velocity form:
Velocity: vf2 = v02 + 2a(xf - x0)
Where:
x0 = 0 m
xf = 100 m
v0 = 0 m/s
vf = 20 m/s
t0 = 0
Thus,
ablue = v2/(2x) = 2.0 m/s2 (slightly less than the acceleration of the red car)
Going back to the position form, we can now calculate how long the blue car is accelerating:
xblue = 1/2 a t2
t = √(2x/a) = 10 s
Blue Car Constant Velocity Phase
We go back to the general form of x(t) to get the expression for the constant velocity phase for the blue car.
Position: x(t) = 1/2 a (t-t0)2 + v0(t-t0) + x0
Where:
x0 = 100 m
v0 = 20 m/s
t0 = 10 s
a = 0
x(t) in meters = 20(t-10) + 100
Thus for the blue car, the full x(t) piecewise expression becomes:
x(t) in meters = t2 ; t < 10 s
x(t) in meters = 20(t-10) + 100 ; t > 10 s
Red Car Acceleration Phase
When does the red car reach top velocity?
vf = v0 + a(t - t0)
Where:
v0 = 0 m/s
vf = 22 m/s
t0 = 3 s
a = 2.5 m/s2
Thus,
t = (vf = v0)/a + t0 = 11.8 s
And how far has the red car gone when it reached top velocity?
x(t) = 1/2 a (t-3)2 = 96.8 m
Thus, the full piecewise x(t) expression for the red car becomes:
x(t) = 0 ; t < 3 s
x(t) = 1.25(t-3)2 ; 3 s < t < 11.8 s
x(t) = 22(t-11.8) + 96.8 ; t > 11.8 s
Now! What do we get?
Well there are a few cases to consider:
Case A: 0 < t < 3 s (trivial - red car not moving)
Case B: 3 < t < 10 s (both cars accelerating)
Case C: 10 s < t < 11.8 s (blue car at constant velocity, red car still accelerating)
Case D: 11.8 s < t (both cars at constant velocity though different velocities)
By setting the appropriate x(t) expressions equal to each other for the cases B, C, D, you then can see if a consistent solution can be found.
For Case B:
t2 = 1.25(t - 3)2
But, solving for t yields no solution within the 3 to 10 s window.
For Case C:
20(t - 10) + 100 = 1.25(t - 3)2
But, solving for t yields no solution within the 10 to 11.8 s window.
For Case D:
20(t - 10) + 100 = 22(t - 11.8) + 96.8
This yields a solution of:
T = 31.4 sec
X = 528 m
I sure hope this is correct after all that. :)
Jeff
