what was the total displacement of the trip?

Question

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


1. Start from rest with constant acceleration of 2.77 m/s2 for 15.0 s.
2. Maintain a constant velocity for the next 2.05 min.
3. Apply a constant negative acceleration of -9.47 m/s2 for 4.39 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?

i would appreciate an explanation along with the calculations. thank you soo much.

Answer 1

1. s1 = 1/2 a1 t1^2; a1 = 2.77 m/sec^2 and t1 = 15 sec
2. s2 = v2 t2 = (a1 t1)t2; where a1 = 2.77 m/sec^2, t1 = 15 sec, and t2 = 2.05 min = 2.05 X 60 sec
3. s3 = 1/2 a3 t3^2; a3 = 9.47 m/sec^2 and t3 = 4.39 sec

a) S = s1 + s2 + s3.
b) S/(t1 + t2 + t3) = V avg for the whole trip; s1/t1 = v1 avg for leg 1, s2/t2 = v2 for leg 2, s3/t3 = v3 for leg 3.

Make sure you do all the times in seconds; so that all the velocities are in m/sec. You can do the arithmetic.