acceleration/velocity problem?

Question

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


1. Start from rest with constant acceleration of 2.27 m/s2 for 16.0 s.
2. Maintain a constant velocity for the next 1.75 min.
3. Apply a constant negative acceleration of -9.82 m/s2 for 3.70 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?

Answer 1

v=u+at
v=0+2.27x16
v=36.32
displacement=(.5x16x36.32)+(1.75x60x36.32)+(.5x3.7x36.32)
=290.56+3813.6+67.192
=4171.352m
Avg speed= total dis/total time=4171.352/(16+1.75x60+37)
=33.45m/s

Answer 2

(a) Total displacement is (1)displacement+(2)displacement+(3)displacement
=0.5*2.27*16^2+[0+2.27*16]*105+[2.27*16]^2/2(9.82)
=290.56+3813.6+67.166=4171.33m

(b) The average velocity in the case of (1) is just the final velocity reached minus the initial velocity of 0 divided by 2, i.e. 2.27*16/2
=18.16m/s. In the case of the second section, the average velocity is just the constant velocity at which the car was travelling, i.e. 2.27*16=36.32m/s. In the third section the average velocity is the same as in the first section since now the initial and final velocities are switched, and so ave. velocity is 18.16m/s.
The average velocity for the whole trip is the complete distance covered divided the time elapsed, i.e. 4171.33m/(16s+105s+3.7s)
=33.45m/s.