Question

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

Answer 1

Answer:

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

$s_1 = a_1t_1^2/2 = 2.2*13^2/2 = 185.9 m$

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

$v_2 = a_1t_1 = 2.2*13 = 28.6 m/s$

The distance traveled for leg 2 is

$s_2 = v_2 t_2 = 28.6*96 = 2745.6 m$

The distance traveled for leg 3 is

$s_3 = v_2t_3 + a_3t_3^2/2 = 28.6*3.12 - 9.16*3.12^2/2 = 44.65 m$

So the total displacement for the trip is

$s = s_1 + s_2 + s_3 = 185.9 + 2745.6 + 44.65 = 2976.15 m$

b) The average speed of leg 1 is

$v_1 = s_1/t_1 = 185.9 / 13 = 14.3m/s$

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

$v_3 = s_3 / t_3 = 44.65 / 3.12 = 14.3104 m/s$

The total time of the entire trip

$t = t_1 + t_2 + t_3 = 13 + 96 + 3.12 = 112.12 s$

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