Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates uniformly at a rate of 3 m/s2 for 3.6 seconds. It then continues at a constant speed for 11.7 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 173.79 meters from where it started. The yellow car accelerates uniformly for the entire distance, finally catching the blue car just as the blue car comes to a stop.

Question

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates uniformly at a rate of 3 m/s2 for 3.6 seconds. It then continues at a constant speed for 11.7 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 173.79 meters from where it started. The yellow car accelerates uniformly for the entire distance, finally catching the blue car just as the blue car comes to a stop.

Answer 1

Wondering what the question is

Answer 2

To solve this problem, we can break it down into smaller parts and use equations of motion to calculate the necessary quantities. Let's start by finding the initial velocity, final velocity, and the distance traveled by the blue car during each phase.

Phase 1: Acceleration
We are given that the blue car accelerates uniformly at a rate of 3 m/s^2 for 3.6 seconds. To find the initial velocity (v0), final velocity (v), and distance traveled (d1) during this phase, we can use the following equations:

v = v0 + at, where v0 = 0 (since the car starts from rest)
d1 = v0t + (1/2)at^2

Using these equations, we can calculate:
v = 3 m/s^2 * 3.6 s = 10.8 m/s (final velocity)
d1 = 0 + (1/2) * 3 m/s^2 * (3.6 s)^2 = 19.44 m (distance traveled)

Phase 2: Constant Speed
We are given that the blue car continues at a constant speed for 11.7 seconds. Since the speed is constant, we know that the acceleration is zero. So, the velocity remains at v = 10.8 m/s during this phase. To find the distance traveled during this phase (d2), we can use the equation:

d2 = vt

Substituting the values, we get:
d2 = 10.8 m/s * 11.7 s = 126.36 m (distance traveled)

Phase 3: Deceleration
We are given that the blue car applies the brakes uniformly and comes to rest 173.79 meters from where it started. To find the deceleration during this phase, we can use the equation:

v^2 = v0^2 + 2ad

Since the final velocity (v) is 0 m/s and the initial velocity (v0) is 10.8 m/s, and the distance (d3) is given as 173.79 m, we can rearrange the equation to solve for the deceleration (a):

0^2 = (10.8 m/s)^2 + 2a * 173.79 m

Solving for a, we find:
a = - (10.8 m/s)^2 / (2 * 173.79 m) = - 0.338 m/s^2 (negative sign indicates deceleration)

Now, let's find the distance traveled (d3) during this phase using the equation:

d3 = (v0^2 - v^2) / (2 * |a|)

Substituting the values, we get:
d3 = (10.8 m/s)^2 / (2 * 0.338 m/s^2) = 173.79 m (distance traveled)

Since the yellow car catches the blue car just as it comes to a stop, this means that the yellow car must have traveled the same distance as the blue car during each phase.

Therefore, the yellow car also travels 19.44 m during phase 1, 126.36 m during phase 2, and 173.79 m during phase 3.

In summary, the distances traveled by the blue car and the yellow car during each phase are as follows:

Blue Car:
Phase 1: 19.44 m
Phase 2: 126.36 m
Phase 3: 173.79 m

Yellow Car:
Phase 1: 19.44 m
Phase 2: 126.36 m
Phase 3: 173.79 m