College Physics Problem?

Question

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

1. Start from rest with constant acceleration of 2.16 m/s2 for 15.0 s.
2. Maintain a constant velocity for the next 1.45 min.
3. Apply a constant negative acceleration of -8.74 m/s2 for 3.71 s.

What was the total displacement for the trip (in meters)?

What were the average speeds for legs 1, 2, and 3 of the trip, as well as for the complete trip (in m/s)?

Please help if you know anything about Physics!

Answer 1

It starts at rest. So x0=0 and v0=0 at the begining.
1) remember that x=x0+v0*t+(1/2)*a*t^2
x1=0.5*2.16m/s2*(15.0s)^2
x1=243m
v(average1)=x1/t1
=243m / 15s
=16.2m/s

2) At the beginnig of 2th leg, the velocity is
v2=a1*t1
=2.16m/s*15s
=32.4m/s
this is also the constant velocity during the second leg.
x2=v2*t2
=32.4m/s*(1.45*60s)
=2818.8m

v(average2)=x2/t2
=2818.8m/(1.45*60s)
=32.4m/s
Notice that velocity is constant during this trip.

3)
x3=v2*t3-(1/2)*a3*(t3)^2
=32.4m/s*(3.71s)- (1/2)*8.74m/s2*(3.71s)^2
=60.055m
v(average3)=x3/t3
=60.055m/3.71s
=16.187m/s
As a result,
Total displacement is
x=x1+x2+x3
=243m+2818.8m+60.055m
=3121.855m
Average speed is
v(average)=x/(t1+t2+t3)
=3121.855m/(15.0s+1.45*60s+3.71s)
=29.532m/s

Answer 2

Calculate the constant velocity:

2.16*15=32.4m/s

now the terminal velocity 32.4-8.74*3.71=-0.0154m/s which is essentially 0.

How far did the mass travel:
32.4(15/2+60*1.45+3.71/2)
=3121.902m

leg 1 is 32.4/2=16.2m/s
Leg 2 is 32.4m/s
leg 3 is 32.4/2=16.2m/s
complete =3121.902/(15+1.45*60+3.71)
=29.53270268m/s

Answer 3

a million) cant answer except you coach the triangular wedge EDIT: ignored relax of that. accel. down slope = 3/5*g = 6 m/s^2 dist = 5 aww, crap, I messsed right here 2) ability potential is in accordance with top. So with a top of zeero, you have 0j of ability potential.