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

a)

b)

c)

d)

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:

And for the acceleration:

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:

And for the acceleration:

the earths atmosphere is made up of nitrogen and oxygen. solids can turn into gasses through a process called sublimation. sublimation is the transition of a substance directly from the solid to the gas phase, without passing through the intermediate liquid phase.

hope this : d

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)