# physics

A record of travel along a straight path is as follows:
1. Start from rest with constant acceleration of 2.59 m/s2 for 19.0 s.
2. Maintain a constant velocity for the next 2.15 min.
3. Apply a constant negative acceleration of −9.08 m/s2 for 5.42 s.
(a) What was the total displacement for the trip?

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1. 1. Vo = a*t = 2.59 * 19 = 49.21 m/s.
d1 = 0.5a*t^2 = 0.5*2.59*19^2 = 467.5 m.

2. d2 = Vo*t = 49.21 * (2.15*60)=6348 m.

3. V = a*t = -9.08 * 5.42 = -49.21 m/s.
V-Vo = 49.21-49.21 = 0 = Final velocity.
d3 = (Vf^2-Vo^2)/2g. Vf = 0
d3 = -(Vo^2)/2g = -(49.21^2)/-19.6 = 124 m.

a. D = d1+d2+d3 = 468 + 6348 + 124 =
6940 m.

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posted by Henry

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