A car starts from rest and travels for 5.0s with a uniform acceleration of +1.5 m/s^2. The driver then applies the brakes, causing a uniform acceleration of -2.0 m/s^2. If the brakes are applied for 3.0 s, how fast is the car going at the end of the braking period, and how far has it gone from its start?
1) find the velocity at the time brakes were applied: Vf=at=7.5m/s
2) then find the final velocity at end of brakeing: Vf=7.5-2*t
1.5 m/s
20m
To solve this problem, we can break it down into two parts: the initial acceleration phase and the braking phase.
1. Initial acceleration phase:
Given:
- Time (t1) = 5.0 s
- Acceleration (a1) = +1.5 m/s^2
We can use the equation of motion to find the final velocity (v1) at the end of this phase.
v1 = u + a1 * t1
Since the car starts from rest, the initial velocity (u) is 0.
v1 = 0 + 1.5 * 5.0
v1 = 7.5 m/s
2. Braking phase:
Given:
- Time (t2) = 3.0 s
- Acceleration (a2) = -2.0 m/s^2
We can again use the equation of motion to find the final velocity (v2) at the end of this phase.
v2 = u + a2 * t2
Since the car ends at v1 (the velocity at the end of the initial phase), the initial velocity for this phase is v1.
v2 = 7.5 + (-2.0) * 3.0
v2 = 7.5 - 6.0
v2 = 1.5 m/s
To find the distance traveled during each phase, we can use the kinematic equation:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
For the initial acceleration phase:
distance1 = (0 * 5.0) + (0.5 * 1.5 * 5.0^2)
distance1 = 0 + (0.5 * 1.5 * 25.0)
distance1 = 0 + 18.75
distance1 = 18.75 m
For the braking phase:
distance2 = (7.5 * 3.0) + (0.5 * (-2.0) * 3.0^2)
distance2 = (7.5 * 3.0) + (0.5 * (-2.0) * 9.0)
distance2 = 22.5 - 9.0
distance2 = 13.5 m
Therefore, at the end of the braking period, the car is going at a speed of 1.5 m/s, and it has traveled a total distance of 18.75 + 13.5 = 32.25 m from its start.
To solve this problem, we can break it down into two parts: the first part where the car accelerates, and the second part where it decelerates.
Part 1: Acceleration
We are given:
Initial velocity (u) = 0 m/s
Acceleration (a1) = +1.5 m/s^2
Time (t1) = 5.0 s
Using the formula:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can calculate the final velocity at the end of the acceleration period.
v1 = 0 + (1.5 m/s^2)(5.0 s)
v1 = 7.5 m/s
So, at the end of the acceleration period, the car is traveling at a speed of 7.5 m/s.
To find the distance traveled during this time, we can use the formula:
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.
s1 = (0 m/s)(5.0 s) + (1/2)(1.5 m/s^2)(5.0 s)^2
s1 = 18.75 m
Therefore, the distance traveled during the acceleration period is 18.75 meters.
Part 2: Deceleration
We are given:
Initial velocity (u2) = 7.5 m/s
Acceleration (a2) = -2.0 m/s^2
Time (t2) = 3.0 s
To find the final velocity at the end of the deceleration period, we use the same formula as before:
v2 = u2 + a2t2
v2 = 7.5 m/s + (-2.0 m/s^2)(3.0 s)
v2 = 7.5 m/s - 6.0 m/s
v2 = 1.5 m/s
So, at the end of the braking period, the car is traveling at a speed of 1.5 m/s.
To find the distance traveled during the deceleration period, we can use the same formula as before:
s2 = (7.5 m/s)(3.0 s) + (1/2)(-2.0 m/s^2)(3.0 s)^2
s2 = 22.5 m - 9.0 m
s2 = 13.5 m
Therefore, the distance traveled during the deceleration period is 13.5 meters.
To find the total distance traveled, we add the distances traveled during the acceleration and deceleration periods:
Total distance = s1 + s2
Total distance = 18.75 m + 13.5 m
Total distance = 32.25 m
Therefore, the car has traveled a total distance of 32.25 meters.
And the final speed of the car at the end of the braking period is 1.5 m/s.