Two cars are approaching one another in the same lane of a highway. One has a speed of 30m/s, the other a speed of 20m/s when they are 96m apart. The drivers simultaneously apply brakes to avert a collision. Both cars lose speed at 6.0m/s^2. How long after the application of brakes do the cars collide? Where are the cars and at what speeds are they travelling when they collide?

Question

Two cars are approaching one another in the same lane of a highway. One has a speed of 30m/s, the other a speed of 20m/s when they are 96m apart. The drivers simultaneously apply brakes to avert a collision. Both cars lose speed at 6.0m/s^2. How long after the application of brakes do the cars collide? Where are the cars and at what speeds are they travelling when they collide?

dA = v1t + 1/2 at^2
= 30t - 3t^2

dB = v1t + 1/2 t^2
= 20t - 3t^2

dA+dB = 96
30t - 3t^2 + 20t - 3t^2 = 96
50t - 6t^2 = 96
- 6t^2 + 50t - 96 = 0
*dividing both sides by -2*
3t^2 - 25t + 48
Should I factor with the quadratic formula?

And then I have no idea how to answer this question -
Where are the cars and at what speeds are they travelling when they collide?

Answer 1

You have it!

Solve the quadratic. You will get 2 answers. One of them, the lower, is the impact time. The other is a "phantom event" where again the distance 96m when the cars back up because the negative acceleration. That event, of course, does not happen.

Answer 2

To solve the equation 3t^2 - 25t + 48 = 0, you can either use factoring or the quadratic formula. Let's use factoring to find the values of t when the cars collide.

To factor the equation 3t^2 - 25t + 48 = 0, we need to find two numbers whose product is 48 and whose sum is -25. The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48.

Since the middle term is negative, the signs of the factors need to be negative in order to get a negative sum. By trying different combinations, we find that -4 and -12 satisfy these conditions.

So, we can rewrite the equation as:

3t^2 - 25t + 48 = (t - 4)(3t - 12) = 0

To find the values of t, we set each factor equal to zero:

t - 4 = 0
t = 4

3t - 12 = 0
t = 12/3
t = 4

Therefore, the cars collide at t = 4 seconds after applying the brakes.

To find the positions and speeds of the cars at the time of collision, we substitute t = 4 into the distance equations:

For car A:
dA = 30t - 3t^2
dA = 30(4) - 3(4)^2
dA = 120 - 3(16)
dA = 120 - 48
dA = 72 meters

For car B:
dB = 20t - 3t^2
dB = 20(4) - 3(4)^2
dB = 80 - 3(16)
dB = 80 - 48
dB = 32 meters

So, at the time of collision, car A is 72 meters from the starting point, car B is 32 meters from the starting point, and they are traveling at speeds of 30 m/s and 20 m/s, respectively.