From t = 0 to t = 3.30 min, a man stands still, and from t = 3.30 min to t = 6.60 min, he walks briskly in a straight line at a constant speed of 1.76 m/s. What are (a) his average velocity vavg and (b) his average acceleration aavg in the time interval 1.00 min to 4.30 min? What are (c) vavg and (d) aavg in the time interval 2.00 min to 5.30 min?
To find the average velocity and average acceleration in a given time interval, we need to calculate the displacement and change in velocity during that interval.
(a) To find the average velocity, we divide the displacement by the time interval.
- In the time interval 1.00 min to 4.30 min, the man is standing still, so his displacement is 0.
- The time interval is 4.30 min - 1.00 min = 3.30 min.
- Therefore, the average velocity is 0 m/3.30 min = 0 m/s.
(b) To find the average acceleration, we divide the change in velocity by the time interval.
- Since the man is standing still, his initial velocity is 0 m/s.
- At t = 4.30 min, the man starts walking briskly at a constant speed of 1.76 m/s.
- Therefore, the change in velocity is 1.76 m/s - 0 m/s = 1.76 m/s.
- The time interval is 4.30 min - 1.00 min = 3.30 min.
- Therefore, the average acceleration is (1.76 m/s) / (3.30 min) = 0.533 m/s².
(c) In the time interval 2.00 min to 5.30 min, the man is walking briskly in a straight line at a constant speed of 1.76 m/s.
- The change in velocity is 1.76 m/s - 1.76 m/s = 0 m/s since the velocity remains constant.
- The time interval is 5.30 min - 2.00 min = 3.30 min.
- Therefore, the average velocity is 0 m/3.30 min = 0 m/s.
- The average acceleration is also 0 m/s² since there is no change in velocity.
(d) Similarly, in the time interval 2.00 min to 5.30 min, the man's change in velocity is 0 m/s and the time interval is 3.30 min.
- Therefore, the average velocity and average acceleration are both 0 m/s and 0 m/s², respectively.
To answer these questions, we need to calculate the average velocity and average acceleration for the given time intervals.
(a) Average velocity vavg in the time interval 1.00 min to 4.30 min:
- The man was standing still from t = 0 to t = 3.30 min, so his initial velocity v0 is 0 m/s.
- The man walked briskly from t = 3.30 min to t = 4.30 min at a constant speed of 1.76 m/s.
- The time interval is 4.30 min - 1.00 min = 3.30 min.
To calculate the average velocity, we use the formula:
vavg = (vf - v0) / (tf - t0)
Substituting the values:
vavg = (1.76 m/s - 0 m/s) / (4.30 min - 1.00 min)
vavg = 1.76 m/s / 3.30 min
vavg = 0.533 m/s
Therefore, the average velocity vavg in the time interval 1.00 min to 4.30 min is 0.533 m/s.
(b) Average acceleration aavg in the time interval 1.00 min to 4.30 min:
- The man's initial velocity v0 is 0 m/s.
- The final velocity vf is 1.76 m/s.
- The time interval is 4.30 min - 1.00 min = 3.30 min.
To calculate the average acceleration, we use the formula:
aavg = (vf - v0) / (tf - t0)
Substituting the values:
aavg = (1.76 m/s - 0 m/s) / (4.30 min - 1.00 min)
aavg = 1.76 m/s / 3.30 min
aavg = 0.533 m/s^2
Therefore, the average acceleration aavg in the time interval 1.00 min to 4.30 min is 0.533 m/s^2.
(c) Average velocity vavg in the time interval 2.00 min to 5.30 min:
- The man was standing still from t = 0 to t = 3.30 min, so his initial velocity v0 is 0 m/s.
- The man walked briskly from t = 3.30 min to t = 5.30 min at a constant speed of 1.76 m/s.
- The time interval is 5.30 min - 2.00 min = 3.30 min.
Using the same formula as in (a), we can calculate the average velocity:
vavg = (vf - v0) / (tf - t0)
vavg = (1.76 m/s - 0 m/s) / (5.30 min - 2.00 min)
vavg = 1.76 m/s / 3.30 min
vavg = 0.533 m/s
Therefore, the average velocity vavg in the time interval 2.00 min to 5.30 min is 0.533 m/s.
(d) Average acceleration aavg in the time interval 2.00 min to 5.30 min:
- The man's initial velocity v0 is 0 m/s.
- The final velocity vf is 1.76 m/s.
- The time interval is 5.30 min - 2.00 min = 3.30 min.
Using the same formula as in (b), we can calculate the average acceleration:
aavg = (vf - v0) / (tf - t0)
aavg = (1.76 m/s - 0 m/s) / (5.30 min - 2.00 min)
aavg = 1.76 m/s / 3.30 min
aavg = 0.533 m/s^2
Therefore, the average acceleration aavg in the time interval 2.00 min to 5.30 min is 0.533 m/s^2.