CG OF BIKE RIDER 30 FT 20 FT 25 FT SolutionAns the initial s

C.G. OF BIKE+ RIDER 3.0 FT 2.0 FT 2.5 FT

Ans

the initial speed=30ft/s

The final velocity =0

The weight of the system is acting at the CG of the system at a height h from the ground.

The brake is applied at the front wheel.

Now the torque due to the weight on the rear wheel = weight X distance X sin(angle)

= 150 lbs X 2.5 ft X sin(90)

= 375 FPS unit

And torque to the retarding force= forceX(height) X sin(angle) = FX 3ft X sin(90)= 3F FPS unit

When these two torques are balanced, the bike comes to rest.

This gives the force on the rear wheel.

So F = 375/3 FPS unit= 125 FPS unit

This force produces acceleration.

hence acceleration(a)= Force/ mass =125/175 f/s²

We will get the distance by using:

v²=u²+2as

Here u=0 .

Hence the required distance will be:

s= v²/2a