Calculating Constant Engine Force for Motorcyclist Acceleration

[chawki]

Homework Statement

Motorcyclist (total mass 320 kg) stops from a velocity of 36km/h in 25 s without breaking.
The same motorcyclist accelerates in the same conditions from full stop to velocity of 126km/h in 6 s.

Homework Equations

Calculate the constant forward carrying force of the engine, if the forces restricting the motion remain constant. Give the answer in 1 N precision.

The Attempt at a Solution

all i could find out is the accelerations, a1=-0.4m/s² and a2= 0.58m/s²
and then i guess we will be interested only in a2, means only the second step of accelerating from a full stop. so F=320*0.58=185.6N
and since they asked to give the answer in 1N precision so it will be 186N.
Please tell me if I'm right

 

[PhanthomJay]

No, you are not right, since you must read the problem carefully. There is a retarding force acting while the cycle is accelerating ...what is that retarding force?

 

[chawki]

What retarding force? the friction force you mean?

 

[PhanthomJay]

What retarding force? the friction force you mean?

Whatever the force is that you calculate in the first part, when the cycle comes to a stop without braking.

 

[chawki]

you mean the friction that brought the cycle to stop? and how do we find that?

 

[PhanthomJay]

It might be friction, or perhaps the cycle is moving uphill...or a combination thereof.. it doesn't matter...you can calculate that retarding force the same way you tried to calculate the motor force in part 2 ---find the acceleration and the retarding force in part 1 using kinematics and Newton 2.

 

[chawki]

Ok well!
F_R is a resistive force.

-F_R=-m*a₁
a₁=0-10/25-0
a₁=-0.4m/s²

-F_R=-m*a₁
F_R=-320*(-0.4)=128N

During acceleration from a full stop, we have:
F-F_R=m*a₂

a₂=35/6=5.83m/s₂

F=m*a₂+F_R
F=(320*5.83)+128
F=1993.6N
and as we have been asked to give the answer in 1N precision, then F=1994N?

 

[PhanthomJay]

Yes, correct.