Q:
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 5.7 m/s2 for 3.2 seconds. It then continues at a constant speed for 8.8 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 218.88 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 1 seconds after it starts?
2)How fast is the blue car going 9.6 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?
7)Below is some space to write notes on this problem
Thank you!
1. V = Vo + a*t = 0 + 5.7*1 = 5.7 m/s.
2. V = Vo + a*t = 0 + 5.7*3.2 = 18.24 m/s. after 3.2 s.
V = 18.24 m/s after 9.6 s.
3. d1 = 0.5a*t^2 = 0.5*5.7*3.2^2 = 29.2 m.
d2 = V*T = 18.24 * 8.8 = 160.5 m.
d = d1+d2 = 29.2 + 160.5 = 189.7 m.
4. V^2 = Vo^2 + 2a*d.
0 = 18.24^2 + 2a*218.9.
a = ?. It should be negative.
5. t1 = 3.2s, t2 = 8.8 s,
V = Vo + a*t.
0 = 18.24 + a*T3,
T3 = ?.
Use the value of "a" calculated in #4.
It should be negative.
T = t1+t2+T3 = Time of movement.
6.
6. 0.5a*T^2 = 189.7m.
a = ?.
T = t1+t2+T3.
To answer these questions, we need to analyze the motion of the blue car step by step. Let's break down the problem:
1) How fast is the blue car going 1 second after it starts?
To find the velocity of the blue car, we need to calculate the acceleration during the first 1 second. Since the blue car is accelerating uniformly, we can use the formula:
v = u + at
Here, v represents the final velocity, u is the initial velocity (which is 0 since the car started from rest), a is the acceleration, and t is the time. Plugging in the given values, we have:
v = 0 + (5.7 m/s^2) * 1s = 5.7 m/s
Therefore, the blue car is going 5.7 m/s after 1 second.
2) How fast is the blue car going 9.6 seconds after it starts?
To find the velocity 9.6 seconds after the car starts, we need to calculate the total displacement during the first 1 second and then during the constant speed phase. For the constant speed phase, we know that velocity is constant, so we can use the formula:
s = vt
Here, s represents displacement, v is the velocity, and t is the time. The displacement during the constant speed phase can be calculated as:
s = (5.7 m/s) * 8.8s = 50.16 m
Adding the displacement during the first 1 second (which we calculated in the previous question):
s = 50.16 m + 5.7 m = 55.86 m
Now, to find the final velocity, we can rearrange the formula:
v = s / t
v = 55.86 m / 9.6 s ≈ 5.814 m/s
Therefore, the blue car is going approximately 5.814 m/s after 9.6 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 calculate the total displacement during the constant speed phase. We have already calculated this displacement in the previous question:
s = 50.16 m
Therefore, the blue car travels a distance of 50.16 meters before applying the brakes.
4) What is the acceleration of the blue car once the brakes are applied?
Since the blue car comes to rest while braking, we know that its final velocity is 0. We can use the following formula to find the acceleration:
v^2 = u^2 + 2as
Here, v represents the final velocity (0 m/s), u is the initial velocity (which is the constant speed achieved during the previous phase), a is the acceleration, and s is the displacement (218.88 m).
0^2 = (5.814 m/s)^2 + 2(a) * 218.88 m
Rearranging and solving for the acceleration, we get:
a = -((5.814 m/s)^2) / (2 * 218.88 m) ≈ -0.0762 m/s^2
Therefore, the acceleration of the blue car once the brakes are applied is approximately -0.0762 m/s^2.
5) What is the total time the blue car is moving?
To find the total time the blue car is moving, we add up the time taken during the first phase of acceleration, the constant speed phase, and the deceleration phase. Let's calculate it:
Total Time = 3.2 s (acceleration) + 8.8 s (constant speed) + t (deceleration)
We know the total displacement before braking is 218.88 m, so we can calculate the deceleration time using the formula:
s = ut + (1/2) * a * t^2
Rearranging the formula, we find:
218.88 m = 0 * t + (1/2) * (-0.0762 m/s^2) * t^2
This is a quadratic equation, and solving it gives us two possible values of t: t = 32 seconds or t = -32 seconds. Since time cannot be negative, the blue car decelerates for 32 seconds.
Plugging in the values into the total time formula:
Total Time = 3.2 s + 8.8 s + 32 s = 44 s
Therefore, the total time the blue car is moving is 44 seconds.
6) What is the acceleration of the yellow car?
Since the yellow car accelerates uniformly for the entire distance and catches up with the blue car just as it comes to a stop, we can assume that the yellow car and the blue car cover the same distance during the constant speed phase.
Therefore, the acceleration of the yellow car is the same as the acceleration of the blue car during the constant speed phase, which is 0 m/s^2. The yellow car is not accelerating or decelerating, but maintaining a constant speed.
7) Below are some notes for this problem:
- The blue car's velocity after acceleration can be calculated using v = u + at, where u is the initial velocity (0 m/s), a is the acceleration (5.7 m/s^2), and t is the time.
- The blue car's displacement during the constant speed phase can be calculated using s = vt, where v is the velocity (5.7 m/s), and t is the time (8.8 seconds).
- The total displacement before braking can be found by adding the displacement during the constant speed phase to the displacement during the first 1 second.
- The acceleration of the blue car once the brakes are applied can be found using the equation v^2 = u^2 + 2as, assuming the final velocity is 0.
- The total time the blue car is moving can be calculated by summing the time taken during the first phase of acceleration, the constant speed phase, and the deceleration phase.
- The acceleration of the yellow car is 0 m/s^2 since it maintains a constant speed.