A motorbike starts from rest at time t=0 and begins to accelerate around a circular track as shown in the figure below. Eventually, at time t=t_1 the motorbike reaches the maximum velocity possible without slipping off the track. What's the minimum length in meters the motorbike must travel between t=0 and t=t_1?

answer please..............

hard

opps, i forgot to write the radius of track, r=10 meters

answer please

To find the minimum length the motorbike must travel between t=0 and t=t_1, we need to consider the factors that determine the maximum velocity without slipping off the track.

First, we need to understand the concept of centripetal acceleration. Centripetal acceleration is the acceleration directed towards the center of a circular path. In this case, the motorbike is accelerating in a circular track, so there must be a centripetal force acting on it.

The centripetal force is provided by the friction between the motorbike's tires and the track. The maximum friction force that can be exerted without slipping can be calculated using the equation:

f_max = µ*N

where f_max is the maximum friction force, µ is the coefficient of friction between the tires and the track, and N is the normal force exerted by the track on the motorbike.

Since the motorbike is at maximum velocity without slipping off the track, the centripetal force is equal to the maximum friction force:

m*a_c = f_max

where m is the mass of the motorbike and a_c is the centripetal acceleration.

The centripetal acceleration can be calculated using the equation:

a_c = v^2 / r

where v is the velocity of the motorbike and r is the radius of the circular track.

Combining these equations, we can find the maximum velocity without slipping:

m * (v^2 / r) = µ * N

We can rearrange this equation to solve for v:

v^2 = (µ * N * r) / m

Now, to find the minimum length the motorbike must travel, we need to calculate the distance traveled during the time interval between t=0 and t=t_1.

We can use the equation for distance traveled during uniform acceleration:

s = ut + (1/2)at^2

In this case, the motorbike starts from rest (u=0) and accelerates uniformly.

The initial velocity (u) is zero, and the final velocity (v) is the maximum velocity without slipping. Therefore, the equation simplifies to:

s = (1/2)at^2

where a is the centripetal acceleration and t is the time taken for the motorbike to reach the maximum velocity without slipping (t_1).

Since we want the minimum length, we need to find the distance traveled at t=t_1.

Now we have all the necessary equations to solve the problem. You will need to have the values of µ (coefficient of friction), m (mass of the motorbike), N (normal force exerted by the track on the motorbike), r (radius of the circular track), and t_1 (time taken to reach the maximum velocity without slipping). Once you have these values, plug them into the equations to find the maximum velocity (v) and the minimum length traveled (s) between t=0 and t=t_1.