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 3.3 m/s2 for 3.9 seconds. It then continues at a constant speed for 13.7 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 220.0 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.
1. How fast is the blue car going 2.0 seconds after it starts?
2. How fast is the blue car going 14.7 seconds after it starts?
3. How far does the blue car travel before its brakes are applied to slow down?
4.What is the acceleration of the blue car once the brakes are applied?
5. What is the total time the blue car is moving?
6. What is the acceleration of the yellow car?
These are busy work problems.
Remember, they both travel the same distance, in the same time.
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To find the answers, we need to analyze the motion of the blue car at different time intervals. We'll break down the problem and solve it step by step.
1. How fast is the blue car going 2.0 seconds after it starts?
To find the velocity at a given time, we can use the equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Since the blue car starts from rest (u = 0) and accelerates uniformly at a rate of 3.3 m/s^2, we can substitute these values into the equation:
v = 0 + (3.3 m/s^2)(2.0 s)
v = 6.6 m/s
So, the blue car is going 6.6 m/s after 2.0 seconds.
2. How fast is the blue car going 14.7 seconds after it starts?
We know that the blue car accelerates for 3.9 seconds, reaches a constant speed, and then decelerates until it comes to rest. To determine the velocity at 14.7 seconds, we need to calculate the distance traveled in each phase.
First, let's find the distance traveled during the acceleration phase using the equation:
s = ut + (1/2)at^2
where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.
For the acceleration phase:
s = (1/2)(3.3 m/s^2)(3.9 s)^2
s = 23.805 m
Next, let's find the distance traveled at the constant speed using the equation:
s = vt
where s is the distance, v is the velocity, and t is the time.
For the constant speed phase, the distance is simply:
s = (6.6 m/s)(13.7 s)
s = 90.42 m
To find the velocity at 14.7 seconds, we'll add the distances traveled during the acceleration and constant speed phases to determine the position of the blue car:
s_total = 23.805 m + 90.42 m
s_total = 114.225 m
Now, to find the velocity at 14.7 seconds, we can rearrange the equation s = vt to solve for v:
v = s/t
v = 114.225 m / 14.7 s
v ≈ 7.775 m/s
So, the blue car is going approximately 7.775 m/s after 14.7 seconds.
3. How far does the blue car travel before its brakes are applied to slow down?
To find the distance traveled before applying the brakes, we need to consider the total distance covered during acceleration and constant speed phases.
From the previous calculations, we found that the distance traveled during the acceleration phase is 23.805 m. The distance traveled at the constant speed is 90.42 m. Therefore, the total distance covered before applying the brakes is:
Total distance = 23.805 m + 90.42 m
Total distance ≈ 114.225 m
So, the blue car travels approximately 114.225 m before the brakes are applied.
4. What is the acceleration of the blue car once the brakes are applied?
Since the blue car slows down uniformly, we can use the same kinematic equation as before:
v = u + at
The final velocity should be 0 m/s since the car comes to rest. The initial velocity will depend on the speed immediately before applying the brakes. From our previous calculation, the speed at 14.7 seconds is approximately 7.775 m/s. The time taken for the brakes can be calculated as follows:
t = (time at 14.7 seconds) - (time during constant speed phase)
t = 14.7 s - 13.7 s
t = 1 s
Now, let's plug in the values into the kinematic equation:
0 = 7.775 m/s + a(1 s)
a ≈ -7.775 m/s^2
Therefore, the acceleration of the blue car once the brakes are applied is approximately -7.775 m/s^2 in the opposite direction of motion.
5. What is the total time the blue car is moving?
To find the total time the blue car is moving, we need to add up the durations of the different phases.
The blue car accelerates for 3.9 seconds, continues at a constant speed for 13.7 seconds, and decelerates until it comes to rest. Therefore, the total time the blue car is moving is:
Total time = 3.9 s + 13.7 s = 17.6 s
So, the blue car is moving for a total of 17.6 seconds.
6. What is the acceleration of the yellow car?
From the problem statement, we know that the yellow car accelerates uniformly for the entire distance and catches up to the blue car just as it comes to a stop. This means that the yellow car covers the same distance in the same time.
The distance traveled by the blue car before the brakes are applied is approximately 114.225 m, and the total time the blue car is moving is 17.6 s.
Using the equation for acceleration:
a = (2s) / t^2
where a is the acceleration, s is the distance, and t is the time.
a = (2 * 114.225 m) / (17.6 s)^2
a ≈ 1.034 m/s^2
Therefore, the acceleration of the yellow car is approximately 1.034 m/s^2.