# Braking then speeding

## 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

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PhanthomJay
Homework Helper
Gold Member
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?

What retarding force? the friction force you mean?

PhanthomJay
Homework Helper
Gold Member
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.

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

PhanthomJay
Homework Helper
Gold Member
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.

Ok well!
FR is a resistive force.

-FR=-m*a1
a1=0-10/25-0
a1=-0.4m/s2

-FR=-m*a1
FR=-320*(-0.4)=128N

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

a2=35/6=5.83m/s2

F=m*a2+FR
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. 