A record of travel along a straight path is as follows: 1. Start from rest with constant acceleration of 2.20 m/s2 for 13.0 s. 2. Maintain a constant velocity for the next 1.60 min. 3. Apply a constant negative acceleration of −9.16 m/s2 for 3.12 s.(a) What was the total displacement for the trip?
m

(b) What were the average speeds for legs 1, 2, and 3 of the trip, as well as for the complete trip?

leg 1 m/s
leg 2 m/s
leg 3 m/s
complete trip m/s

a) s = 2976.15 m

b) v1 = 14.3 m/s

v2 = 28.6 m/s

v3 = 14.3104 m/s

v = 26.54 m/s

Explanation:

1.6 min = 1.6 * 60 = 96 s

(a) The distance traveled for leg 1

The speed at the end of leg 1 and for all of leg 2 is

The distance traveled for leg 2 is

The distance traveled for leg 3 is

So the total displacement for the trip is

b) The average speed of leg 1 is

The average speed of leg 2 is the same as the constant speed, which is 28.6 m/s

The average speed of leg 3 is

The total time of the entire trip

So average speed of the entire trip is the total distance over total time

v = s / t = 2976.15 / 112.12 = 26.54 m/s

f=ma newton's second law 7x8=56n