(a) To find how fast the car is going at the end of the braking period, we first need to find its speed before applying the brakes.
We can use the formula vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.
For the first part of the journey (before applying the brakes), we have:
vf1 = 0 + (1.2 m/s²)(6.0 s)
vf1 = 7.2 m/s
Now, we need to find the velocity at the end of the braking period. For this part of the journey, we have:
vi = vf1 = 7.2 m/s
a = -1.8 m/s² (because it's decelerating)
t = 1.70 s
Using the same formula as before:
vf2 = vi + at
vf2 = 7.2 m/s + (-1.8 m/s²)(1.70 s)
vf2 = 7.2 m/s - 3.06 m/s
vf2 ≈ 4.14 m/s
The final velocity at the end of the braking period is approximately 4.14 m/s.
(b) To find the total distance traveled, we first need to find the distance traveled during each part of the journey.
For the first part, we can use the formula d = vi * t + 0.5 * a * t², where d is the distance traveled, vi is the initial velocity, a is the acceleration, and t is the time.
d1 = 0 * 6.0 s + 0.5 * (1.2 m/s²) * (6.0 s)²
d1 = 0 + 0.5 * 1.2 m/s² * 36 s²
d1 = 0 + 0.6 m/s² * 36 s²
d1 = 21.6 m
For the second part, we have:
vi = 7.2 m/s
a = -1.8 m/s²
t = 1.70 s
d2 = vi * t + 0.5 * a * t²
d2 = 7.2 m/s * 1.70 s + 0.5 * (-1.8 m/s²) * (1.70 s)²
d2 = 12.24 m - 0.9 m/s² * 2.89 s²
d2 = 12.24 m - 2.601 m
d2 ≈ 9.639 m
Now, we can find the total distance traveled:
d_total = d1 + d2
d_total = 21.6 m + 9.639 m
d_total ≈ 31.239 m
The car has traveled approximately 31.239 meters from its start.
A car starts from rest and travels for 6.0 s with a uniform acceleration of +1.2 m/s2. The driver then applies the brakes, causing a uniform acceleration of -1.8 m/s2. The breaks are applied for 1.70 s. (a) How fast is the car going at the end of the braking period?m/s (b) How far has the car gone from its start? .
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