SOLUTION: PLEASE HELP!!!!!!!!!! Motorcyclist with a total mass 320 kg stops from a velocity of 36km/h in 25 s without braking. The same motorcyclist accelerates in the same conditions fro

Algebra ->  Testmodule -> SOLUTION: PLEASE HELP!!!!!!!!!! Motorcyclist with a total mass 320 kg stops from a velocity of 36km/h in 25 s without braking. The same motorcyclist accelerates in the same conditions fro      Log On


   

Question 408134: PLEASE HELP!!!!!!!!!!
Motorcyclist with a total mass 320 kg stops from a velocity of 36km/h in 25 s without braking.
The same motorcyclist accelerates in the same conditions from full stop to velocity of 126km/h in 6 s. Calculate the constant forward carrying force of the engine, if the forces restricting the motion remain constant. Give the answer in 1 N precision
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

You can calculate the constant force restricting motion from the first part of the question, where he slows down without braking.

First convert the given speed to m%2Fs:
u+=+36+%281000%2F3600%29+=+10+m%2Fs}
v+=+0
s+= ?
a+ =?
t+=+25s

v+=+u+%2B+at
0+=+10+%2B+25a
-25a+=+10
a+=+-10%2F25
a+=+-0.4+m%2Fs%5E2

F+=+ma
F+=+320+%28-0.4%29
F+=+-128+N
So there is a force resisting motion with a magnitude of 128+N. Next calculate his acceleration:

u+=+0
v+=+126+%2A+%281000+%2F+3600%29+=+35+m%2Fs
s =?
a=?
t+=+6s

v+=+u+%2B+at
35+=+0+%2B+6a
a+=+35%2F6
a+=+5.833+m%2Fs
Now calculate the force+required to give him this acceleration, remembering that there's a resistive force in play which we'll call R:
R+%2B+F+=+ma
F+=+ma+-+R
F+=+320+%2A+5.833+-+%28-128%29
F+=+1%2C866.667+%2B+128
F+=+1994.667
F+=+1995+N

So the engine supplies a carrying force of 1995+N.