What is the answer to this physics question?

Question

From t = 0 to t = 4.23 min, a man stands still, and from t = 4.23 min to t = 8.46 min, he walks briskly in a straight line at a constant speed of 1.34 m/s. What are (a) his average velocity vavg and (b) his average acceleration aavg in the time interval 1.00 min to 5.23 min? What are (c) vavg and (d) aavg in the time interval 2.00 min to 6.23 min?

Answer 1

a.) Note that 4.23 is half of 8.46 - this makes it pretty simple.
First, define average velocity:
avg velocity = (v1t1 + v2t2)/(t1 + t2)
v1 is the initial velocity
v2 is the second velocity
t1 is the amount of time that the man is at the initial velocity
t2 is the amount of time that the man is at the second velocity

v avg = (0 * 4.23 + 1.34 * 4.23) / 8.46
v avg = 0.67 m/s

b.) This is similar to a.
average acceleration = Δv/Δt
Δv = change in velocity
Δt = time interval

a avg = (1.34 - 0) / (5.23 - 1)
a avg = 1.34 / 4.23 = 0.317 m/s^2

c.) Same as #1, but you have to find t1 and t2 by subtracting the times given from the time that the velocity changes.
v avg = (v1t1 + v2t2)/(t1 + t2)
v avg = (0 * (4.23 - 2) + 1.34 * (6.23 - 4.23) / (6.23 - 2)
v avg = 1.34 * 2 / 4.23
v avg = 0.633 m/s

d.) a avg = Δv/Δt
Δv = 1.34
Δt = 6.23 - 2 = 4.23
a avg = 1.34 / 4.23 = 0.317 m/s^2