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 5.5 m/s2 for 4.9 seconds. It then continues at a constant speed for 6.7 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 292.81 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 8 seconds after it starts?
2) How far does the blue car travel before its brakes are applied to slow down?
3) What is the acceleration of the blue car once the brakes are applied?
4) What is the total time the blue car is moving?
5) What is the acceleration of the yellow car?
This would make an excellent graphic solution...I am certain you have done distance, velocity graphs.
If you want to nerd it out in math,
work it in steps.
1. figure speed at 4.9 seconds, then that is te speed at 8 seconds..
2. how far? figure the distance while accelerating, then the distance at constant speed. add them
3. use Vf^2=Vi^2+2ad solve for d, vi is the speed at brake application.
just work it in steps.
you mom
To solve these questions, we'll break down the information given step by step.
1) To find the speed of the blue car 8 seconds after it starts, we need to consider the three stages: acceleration, constant speed, and deceleration.
- For the first stage of acceleration, the blue car accelerates uniformly at a rate of 5.5 m/s^2 for 4.9 seconds. We can use the equation of motion:
v = u + at,
where:
v = final velocity,
u = initial velocity (which is 0 m/s since the car starts from rest),
a = acceleration (5.5 m/s^2),
t = time (4.9 seconds).
Plugging in the values, we get:
v = 0 + (5.5)(4.9) = 26.95 m/s.
- For the second stage of constant speed, the blue car travels for 6.7 seconds at this velocity, so the speed remains constant at 26.95 m/s.
- For the third stage of deceleration, the blue car slows down uniformly to a stop. The distance covered during this stage is given as 292.81 meters. We can use another equation of motion to find the deceleration:
v^2 = u^2 + 2as,
where:
s = distance (292.81 meters),
a = acceleration (which we need to find),
v = final velocity (0 m/s),
u = initial velocity (26.95 m/s).
Rearranging the equation, we get:
a = (v^2 - u^2) / 2s,
Plugging in the values, we get:
a = (0^2 - 26.95^2) / (2 * 292.81) ≈ -5.22 m/s^2.
Therefore, the blue car's speed 8 seconds after it starts is 26.95 m/s.
2) To find the distance traveled by the blue car before its brakes are applied, we need to calculate the distance covered during the first two stages.
- For the first stage of acceleration, we can use the equation of motion:
s = ut + (1/2)at^2,
where:
s = distance,
u = initial velocity (0 m/s),
a = acceleration (5.5 m/s^2),
t = time (4.9 seconds).
Plugging in the values, we get:
s1 = 0 + (0.5)(5.5)(4.9)^2 ≈ 67.84 meters.
- For the second stage of constant speed, since the velocity remains the same for 6.7 seconds, the distance covered is given by:
s2 = v * t = 26.95 * 6.7 ≈ 180.07 meters.
Therefore, the total distance traveled by the blue car before its brakes are applied is s1 + s2 ≈ 67.84 + 180.07 ≈ 247.91 meters.
3) The acceleration of the blue car once the brakes are applied is calculated using the same equation of motion as earlier:
a = (v^2 - u^2) / 2s,
where:
s = total distance (292.81 meters),
u = initial velocity before deceleration (26.95 m/s),
v = final velocity (0 m/s).
Plugging in the values, we get:
a = (0^2 - 26.95^2) / (2 * 292.81) ≈ -5.22 m/s^2.
Therefore, the acceleration of the blue car once the brakes are applied is approximately -5.22 m/s^2.
4) The total time the blue car is moving consists of the time spent in acceleration and the time spent at a constant speed.
Total time = time for the first stage of acceleration + time for the second stage of constant speed
= 4.9 seconds + 6.7 seconds
= 11.6 seconds.
Therefore, the blue car is moving for a total of 11.6 seconds.
5) The yellow car accelerates uniformly for the entire distance and catches up with the blue car just as the blue car comes to a stop.
Since both cars cover the same distance during the same time (the blue car's deceleration time), we can find the acceleration of the yellow car using the equation of motion:
s = ut + (1/2)at^2,
where:
s = total distance covered (292.81 meters),
u = initial velocity (0 m/s),
a = acceleration of the yellow car (which we need to find),
t = time (6.7 seconds, which is the same as the blue car's constant speed time).
Plugging in the values, we get:
292.81 = 0 + (1/2)a(6.7)^2,
292.81 = (1/2)(44.89)a,
a ≈ 13.03 m/s^2.
Therefore, the acceleration of the yellow car is approximately 13.03 m/s^2.