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 4.6 m/s2 for 4.4 seconds. It then continues at a constant speed for 8.5 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 247 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.
2)How fast is the blue car going 10.4 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?
1. V1 = Vo + at,
V1 = 0 + 4.6*4.4 = 20.24m/s.
2. V2 = V1 = 20.24m/s after 10.4s.
3. D = di + d2,
d1 = Vo*t + 0.5at^2,
d1 = 0 + 0.5*4.6*(4.4)^2 = 44.53m.
d2 = Vt = 20.24 * 8.5 = 172m.
D = 44.53 + 172 = 217m.
4. Ds = 247 - 217 = 30m = Stopping
distance.
a = (Vf^2 - Vo^2) / 2d,
a = (0 -(20.24)^2 / 60 = -6.83m/s^2.
To answer these questions, we need to break down the motion of the blue car into three stages:
1) Acceleration phase (from rest):
Given:
Initial velocity (u) = 0 m/s
Acceleration (a) = 4.6 m/s^2
Time (t1) = 4.4 seconds
Using the formula:
Final velocity (v1) = u + a*t1
Calculating the final velocity during the acceleration phase:
v1 = 0 + 4.6 * 4.4
v1 = 20.24 m/s
2) Constant speed phase:
Given:
Time (t2) = 8.5 seconds
Since the car moves at a constant speed during this phase, the velocity remains the same. Therefore, the velocity at the end of this phase (v2) is equal to v1:
v2 = v1 = 20.24 m/s
3) Deceleration phase (applying brakes):
Given:
Initial velocity (u3) = v2 = 20.24 m/s
Final velocity (v3) = 0 m/s
Distance (s) = 247 meters
Using the formula:
s = (u3 + v3) / 2 * t3
We need to solve for t3:
247 = (20.24 + 0) / 2 * t3
494 = 20.24 * t3
t3 = 24.4 seconds
So, the time taken to stop the blue car is 24.4 seconds.
Now, we can answer the remaining questions:
2) How fast is the blue car going 10.4 seconds after it starts?
During the acceleration phase, the velocity increases uniformly. Therefore, we can use the formula:
v = u + a*t
For t = 10.4 seconds:
v = 0 + 4.6 * 10.4
v = 47.84 m/s
Therefore, the blue car is going 47.84 m/s 10.4 seconds after it starts.
3) How far does the blue car travel before its brakes are applied to slow down?
During the acceleration phase, the distance traveled (s1) is given by the formula:
s1 = u*t + 0.5*a*t^2
Substituting the values:
s1 = 0 + 0.5 * 4.6 * (4.4)^2
s1 = 43.168 meters
During the constant speed phase, the distance traveled (s2) is given by the formula:
s2 = v * t2
s2 = 20.24 * 8.5
s2 = 172.04 meters
So, before applying the brakes, the blue car travels a total distance of:
s_total = s1 + s2
s_total = 43.168 + 172.04
s_total = 215.208 meters
Therefore, the blue car travels 215.208 meters before its brakes are applied to slow down.
4) What is the acceleration of the blue car once the brakes are applied?
Since the car comes to rest from a certain velocity (v2) in a certain time (t3), we can use the formula:
a = (v3 - u3) / t3
Substituting the values:
a = (0 - 20.24) / 24.4
a = -20.24 / 24.4
a ≈ -0.8295 m/s^2
Therefore, the acceleration of the blue car once the brakes are applied is approximately -0.8295 m/s^2.
To solve these questions, we need to break down the motion of the blue car into different phases and calculate the relevant quantities for each phase.
First, let's identify the different phases of the blue car's motion:
Phase 1: Acceleration phase - The blue car accelerates uniformly for 4.4 seconds.
Phase 2: Constant speed phase - The blue car maintains a constant speed for 8.5 seconds.
Phase 3: Deceleration phase - The blue car applies the brakes and comes to rest in 247 meters.
1) To find the speed of the blue car 10.4 seconds after it starts, we need to determine which phase the car is in at that time.
Phase 1 duration = 4.4 seconds
Phase 2 duration = 8.5 seconds
Since 10.4 seconds is greater than the sum of phase 1 and phase 2 durations (10.4 > 4.4 + 8.5), we can conclude that the blue car is in the deceleration phase (phase 3) at this time.
During the deceleration phase, the car is slowing down. Given that the blue car comes to rest in 247 meters, we can say that the car has traveled 247 meters before its brakes are applied to slow down.
2) To calculate the speed of the blue car 10.4 seconds after it starts, we need to first determine the speed at the end of phase 2 (constant speed phase).
During the constant speed phase (phase 2), the blue car maintains a constant speed. To find this speed, we can use the formula:
Speed = Acceleration * Time
The acceleration during phase 1 is given as 4.6 m/s^2, and the duration of phase 1 is 4.4 seconds. Therefore, the speed at the end of phase 1 (beginning of phase 2) is:
Speed at the end of phase 1 = 4.6 m/s^2 * 4.4 s = 20.24 m/s
Since the blue car maintains this speed during phase 2, the speed at the end of phase 2 (beginning of phase 3) is also 20.24 m/s.
Now, we can calculate the speed of the blue car 10.4 seconds after it starts. Since the blue car is in phase 3 at this time (deceleration phase), the speed is decreasing uniformly until it comes to a rest.
To find the speed at 10.4 seconds, we can use the formula:
Final Speed = Initial Speed + (Acceleration * Time)
The initial speed at the beginning of phase 3 is 20.24 m/s, and the deceleration occurs over a time period of 8.5 seconds. Therefore, the speed at 10.4 seconds is:
Final Speed = 20.24 m/s - (Deceleration * (10.4 s - 8.5 s))
(Deceleration is unknown)
3) As mentioned earlier, the distance traveled by the blue car before its brakes are applied to slow down (phase 3) is 247 meters.
4) Finally, to find the acceleration of the blue car once the brakes are applied (phase 3), we can use the formula:
Acceleration = (Final Speed - Initial Speed) / Time
In this case, the initial speed is 20.24 m/s (at the beginning of phase 3), the final speed is 0 m/s (when the car comes to a stop), and the time is 8.5 seconds.
Acceleration = (0 m/s - 20.24 m/s) / 8.5 s