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 4.8 m/s2 for 4.5 seconds. It then continues at a constant speed for 9.9 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 309 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. How fast is the blue car going 2.3 seconds after it starts?

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 4.8 m/s2 for 4.5 seconds. It then continues at a constant speed for 9.9 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 309 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. How fast is the blue car going 2.3 seconds after it starts?

Answer 1

You just use the equation vf=vi+at. Your initial velocity would be zero since it's starting from rest, your acceleration in this case would be 4.8 m/s^2, and your time would be 2.3 seconds. Your answer will be 11.04 m/s.

Answer 2

To find the speed of the blue car 2.3 seconds after it starts, we need to break down the movement of the blue car into different phases and calculate the distances and speeds for each phase.

Phase 1: Acceleration
During this phase, the blue car accelerates uniformly at a rate of 4.8 m/s^2 for 4.5 seconds. To find the distance covered during this phase, we can use the formula:
distance = initial velocity * time + 0.5 * acceleration * time^2

The initial velocity is 0 since the car starts from rest, the time is 4.5 seconds, and the acceleration is 4.8 m/s^2.
distance_phase1 = 0 * 4.5 + 0.5 * 4.8 * (4.5)^2

Phase 2: Constant Speed
During this phase, the blue car travels at a constant speed for 9.9 seconds. To find the distance covered during this phase, we can use the formula:
distance = speed * time

The speed during this phase is constant but unknown, so we will denote it as V.
distance_phase2 = V * 9.9

Phase 3: Deceleration
During this phase, the blue car decelerates uniformly until it comes to rest after traveling a distance of 309 meters. The final speed is 0, and the distance covered is given as 309 meters.

We can find the deceleration using the formula:
final velocity^2 = initial velocity^2 + 2 * acceleration * distance

Since the final velocity is 0 and the initial velocity is V (from phase 2), and the distance is 309 meters, we can solve for the deceleration.

0 = V^2 + 2 * acceleration * 309

Now that we have the deceleration, we can find the time taken to decelerate using the formula:
final velocity = initial velocity + acceleration * time

Since the initial velocity is V (from phase 2) and the final velocity is 0, we can solve for the time taken to decelerate.

0 = V - deceleration * time_deceleration

Lastly, we can calculate the distance covered during the deceleration phase using the formula:
distance = initial velocity * time + 0.5 * deceleration * time^2

The initial velocity is V (from phase 2), the time is time_deceleration (found above), and the deceleration is the calculated deceleration value.

Now, to calculate the speed of the blue car 2.3 seconds after it starts, we need to add up the distances covered during each phase: distance_phase1 + distance_phase2 + distance_deceleration. Then, divide that total distance by the total time taken: 4.5 seconds (phase 1) + 9.9 seconds (phase 2) + time_deceleration (phase 3).

This will give us the average speed over the entire journey, from which we can find the speed 2.3 seconds after starting.