Question

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 1.39 m/s.

(a) what is his average velocity in the time interval 1.00min to 4.30min?

(b) what is his average acceleration in the time interval 1.00min to 4.30min?

(c) what is his average velocity in the time interval 2.00 min to 5.30 min?

(d) what is his average acceleration in the time interval 2.00 min to 5.30 min?

Answer 1

(a)

distance travelled in 1 min, (from t = 3.3 min to t = 4.3)

d = 1.39*1*60 = 83.4 m

From 1.00 min to 4.30 min,

t = 4.30 - 1 = 3.3 min

so,  average velocity in this time interval,

vavg = 83.4 / 3.3

vavg = 25.27 m/min = 0.421 m/s

(b)

From 1.00 min to 4.30 min,

t = 4.30 - 1 = 3.3 min

average acceleration, a = 1.39 / (3.3*60)

a = 0.00702 m/s^2

(c)

From t = 3.3 to t = 5.30

t = 2 min

distance travelled in 2 min,

d = 1.39*60*2 = 166.8 m

average velocity in the time interval 2.00 min to 5.30 min,

vavg = 166.8 / (5.30 - 2)

vavg = 50.54 m / min = 0.842 m/s

(d)

From 2.00 min to 5.30 min,

t = 3.3 min

a = (1.39) / 3.3*60

a = 0.00702 m/s^2