Question

From t = 0 to t = 3.36 min, a man stands still, and from t = 3.36 min to t = 6.72 min, he walks briskly in a straight line at a constant speed of 1.19 m/s. What are (a) his average velocity vavg and (b) his average acceleration aavg in the time interval 1.00 min to 4.36 min? What are (c) vavg and (d) aavg in the time interval 2.00 min to 5.36 min? (a) Number Enter your answer for part (a) in accordance to the question statement Units Choose the answer for part (a) from the menu in accordance to the question statement

Answer 1

a)
`V_(avg) = 21.25m/min`

b)
`a_(avg) = 21.25m/min^(2)`

c)
`V_(avg) = 42.5m/min`

d)
`a_(avg) = 21.25m/min^(2)`

Step-by-step explanation:

Before doing anything, we need the speed in the same units as time:

V = 1.19m/s * 60s/min = 71.4m/min

Now, for part (a) we need position at t=1min and t=4.36min. Position at t=1min was 0m and at t=4.36min:

X=V*Δt = 71.4*(4.36-3.36)=71.4m

With this value, we can proceed to calculate average speed:

`V_(avg)=(X_(4.36)-X_(1))/(4.36-1)=21.25m/min`

And for the acceleration:

`a_(avg)=(V_(4.36)-V_(1))/(4.36-1)=(71.4-0)/(3.36)=21.25m/min^(2)`

For part (c) and (d) we proceed similarly:

X=V*Δt = 71.4*(5.36-3.36)=142.8m

With this value, we can proceed to calculate average speed:

`V_(avg)=(X_(5.36)-X_(2))/(5.36-2)=42.5m/min`

And for the acceleration:

`a_(avg)=(V_(5.36)-V_(2))/(5.36-2)=(71.4-0)/(3.36)=21.25m/min^(2)`