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? .
(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.
To solve this problem, we can use the equations of motion. Let's break it down step-by-step.
Step 1: Find the final velocity after the acceleration phase.
We are given:
Initial velocity (u) = 0 m/s (since the car starts from rest)
Acceleration (a) = +1.2 m/s² (uniform acceleration)
Time (t) = 6.0 s
Using the equation: v = u + at
v = 0 + (1.2 × 6.0)
v = 7.2 m/s
So, at the end of the acceleration phase, the car's velocity is 7.2 m/s.
Step 2: Find the final velocity after the braking phase.
We are given:
Acceleration (a) = -1.8 m/s² (uniform acceleration, negative sign indicates deceleration)
Time (t) = 1.70 s
Using the equation: v = u + at
v = 7.2 + (-1.8 × 1.70)
v = 7.2 - 3.06
v = 4.14 m/s
So, at the end of the braking period, the car's velocity is 4.14 m/s.
(a) Answer: The car is going at a speed of 4.14 m/s at the end of the braking period.
Step 3: Find the distance traveled during the acceleration phase.
Using the equation: s = ut + (1/2)at²
s = 0 × 6.0 + (1/2) × 1.2 × (6.0)²
s = 0 + 0.5 × 1.2 × 36.0
s = 21.6 m
Step 4: Find the distance traveled during the braking phase.
Using the equation: s = ut + (1/2)at²
The initial velocity of the braking phase (u) can be taken as the final velocity of the acceleration phase, which is 7.2 m/s.
s = 7.2 × 1.70 + (1/2) × (-1.8) × (1.70)²
s = 12.24 - 2.448
s = 9.792 m
Step 5: Find the total distance traveled.
Total distance = distance during acceleration phase + distance during braking phase
Total distance = 21.6 + 9.792
Total distance = 31.392 m (rounded to three decimal places)
(b) Answer: The car has traveled a distance of 31.392 meters from its start.
To summarize:
(a) The car is going at a speed of 4.14 m/s at the end of the braking period.
(b) The car has traveled a distance of 31.392 meters from its start.
To find the answers to these questions, we need to use the equations of motion and the principles of kinematics. Let's break it down step by step.
(a) How fast is the car going at the end of the braking period?
We can use the equation:
v = u + at
where v represents the final velocity, u represents the initial velocity, a represents the acceleration, and t represents the time.
In the given scenario, the car starts from rest (u = 0), and the acceleration is -1.8 m/s^2 during the braking period (t = 1.70 s).
Using the equation, we can find the final velocity (v):
v = 0 + (-1.8 m/s^2) * (1.70 s)
v = -3.06 m/s
Therefore, the car is traveling at -3.06 m/s (in the opposite direction of its initial motion) at the end of the braking period.
(b) How far has the car gone from its start?
We can use the equation:
s = ut + (1/2)at^2
where s represents the distance traveled, u represents the initial velocity, a represents the acceleration, and t represents the time.
During the first part of the motion (before applying the brakes), the car experiences a uniform acceleration of +1.2 m/s^2 for a time of 6.0 s.
Using the equation, we can find the distance traveled (s) during the first part:
s = 0 + (1/2)(1.2 m/s^2)(6.0 s)^2
s = 0 + (1/2)(1.2 m/s^2)(36.0 s^2)
s = 0 + (1/2)(43.2 m)
s = 21.6 m
During the second part of the motion (while applying the brakes), the car experiences a uniform acceleration of -1.8 m/s^2 for a time of 1.70 s.
Using the equation, we can find the distance traveled (s) during the second part:
s = 0 + (1/2)(-1.8 m/s^2)(1.70 s)^2
s = 0 + (1/2)(-1.8 m/s^2)(2.89 s^2)
s = 0 + (1/2)(-5.242 m)
s = -2.621 m
Adding up the distances traveled during both parts of the motion, we get:
Total distance traveled = 21.6 m + (-2.621 m) = 18.979 m
Therefore, the car has traveled approximately 18.979 m from its start.