Find the speed of the car, in m/s, as it reaches the red light

Question

The driver of a car traveling at 22.9 m/s sees a yellow light ahead. The car is 37.7 m from the light when the light turns red. At that moment, the driver applies the brakes as hard as possible, and the car slows uniformly at a rate of 4.10 m/s². Find the speed of the car, in m/s, as it reaches the red light. 

Equations 1:

0=V1+at

X2=V1t + 1/2at^2

Answer 1

Broadly speaking, here's how we'll solve this problem:

Step 1: Set up our equation to calculate the time it takes for the car to reach the red light

Step 2: Substitute in any known values and solve our equation for time

Step 3: Calculate the final velocity using time, acceleration, and initial velocity

Now, let's go through the steps one at a time.

Step 1:

We can use the equation d = v₁t + 1/2 at² to calculate the time. Our distance is 37.7m, our acceleration is -4.1m/s² (since we're decelerating), and our initial velocity is 22.9 m/s. Thus, we can substitute these values, and we get 37.7 = 22.9t + 1/2*(-4.1t²).

Step 2:

Now, we just need to solve this equation. Subtracting 37.7 from both sides gets us -2.05t²+22.9t-37.7 = 0. This is a quadratic equation that can be solved with the quadratic formula, with a = -2.05, b = 22.9, and c = -37.7. Using the quadratic formula gets us two solutions, t = 2 and t = 9.164. We're interested in when the vehicle first hits the red light, so we take the lower number, and conclude that t = 2 seconds. Doing a quick "sanity check", we can see that our initial velocity is around half of the distance to the red light, and we aren't decelerating that quickly, so it makes sense that our answer is 2 seconds.

Step 3:

We can calculate the change in velocity using v_f = v_i + at. As we saw earlier, v_i = 22.9 m/s, a = -4.1 m/s², and t = 2 seconds. Plugging this in gets us v_f = 22.9 - 4.1(2) = 14.7 m/s.

Thus, the speed of the car when it reaches the red light is 14.7 m/s.