Please help me solve this question

Question

A record of travel along a straight path is as follows:

1. Start from rest with constant acceleration of 2.00 m/s² for 19.0 s.

2. Maintain a constant velocity for the next 2.30 min.

3. Apply a constant negative acceleration of −9.44 m/s² for 4.03 s.

(a) What was the total displacement for the trip?

 

(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

Answer 1

Average speed = Δx/t

Leg 1:

v_i = 0 m/s

a = 2.00 m/s²

t = 19.0 s

Δx = ?

Use the equation Δx = v_it + (1/2)at²

Δx = (0)(19) + (1/2)(2)(19²) = 0 + 361 = 361 meters

average speed = 361/19 = 19 m/s

Leg 2:

v_i = 38.0 m/s

v_f = 38.0 m/s

average speed = 38.0 m/s

t = 2.30 minutes = (2.3)(60) seconds = 138 seconds

Δx = v_avg•t = (38)(138) = 5244 meters

Leg 3:

vi = 38 m/s

a = -9.44 m/s²

t = 4.03 seconds

Δx = ?

Use equation

Δx = v_it + (1/2)at²

Δx = 38(4.03) + (1/2)(-9.44)(4.03²) = 153.14 - 76.657048 = 76.482952 meters

average speed = 76.482952/4.03 = 19.0 m/s

Total displacement = Δx_{(leg 1)} + Δx_{(leg 2)} + Δx_{(leg 3)} = 361 + 5244 + 76.5 = 5680 meters

Total time = t_(leg1) + t_(leg2) + t_(leg3) = 19 + 138 + 4.03 = 161 seconds

Average speed for the entire trip = 5680/161 = 35.3 m/s