Question

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

1. Start from rest with constant acceleration of 2.04 m/s² for 11.0 s.
2. Maintain a constant velocity for the next 2.85 min.
3. Apply a constant negative acceleration of −9.73 m/s² for 2.31 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

Answer 1

first displacement: x1 = (1/2)at² = (1/2)(2.04 m/s²)(11.0 s)² = 123.42 m (to 3 significant figures)

speed at end of first acceleration: v1 = at = (2.04 m/s²)(11.0 s) = 22.44 m/s

displacement while coasting: x = v1t = (22.44 m/s)(185 s) = 4151.4 m

speed while coasting: v2 = 22.44 m/s

displacement while decelerating: x3 = (1/2)at² = (1/2)(9.73 m/s²)(4.31 s)^2 = 90.37 m (the negative sign of the acceleration does not matter here)

final speed after decelerating: v3 = v2 + at = 22.44 m/s + (-9.43 m/s²)(4.31 s) = -18.20

total displacement = 123.42 m + 4151.4 m + 90.37+(-18.20) m = 4346.99 m

first average speed = (v0+v1)/2 = (-18.20 + 22.44 m/s)/2 = 2.12 m/s

second average speed = 22.44 m/s

third average speed = (v2+v3)/2 = (22.44 + -18.20 m/s)/2 = 2.12 m/s.