Braking then speeding

  • Thread starter chawki
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  • #1
chawki
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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
 

Answers and Replies

  • #2
PhanthomJay
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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?
 
  • #3
chawki
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What retarding force? the friction force you mean?
 
  • #4
PhanthomJay
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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.
 
  • #5
chawki
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you mean the friction that brought the cycle to stop? and how do we find that?
 
  • #6
PhanthomJay
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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.
 
  • #7
chawki
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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?
 
  • #8
PhanthomJay
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Yes, correct.:approve:
 

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