A car is driving on a circular flat tarmac of radius R = 55 m (as shown below), starting from rest when t=0 seconds.The tangential component of the car’s acceleration as a function of time is represented by at = 0.5t m/s2.When t=6 seconds, determine the magnitude of the car's:(a) velocity, v = Answer m/s,(b) acceleration, a = Answer m/s2,(c) displacement, s = Answer metres.

(a) Velocity

The tangential acceleration is given by the equation at = 0.5t. This is the rate of change of velocity, so to find the velocity at a given time, we integrate this equation with respect to time.

∫at dt = ∫0.5t dt = 0.25t^2 + C

Since the car starts from rest, the constant of integration C = 0. So, the velocity at t = 6 seconds is:

v = 0.25*(6)^2 = 9 m/s

(b) Acceleration

The acceleration of the car is the vector sum of the tangential acceleration and the radial (or centripetal) acceleration. The tangential acceleration at t = 6 seconds is at = 0.5*6 = 3 m/s^2. The radial acceleration is given by the equation ar = v^2/R, so at t = 6 seconds, ar = (9)^2/55 = 1.47 m/s^2.

The total acceleration is then the square root of the sum of the squares of the tangential and radial accelerations:

a = sqrt((at)^2 + (ar)^2) = sqrt((3)^2 + (1.47)^2) = 3.36 m/s^2

(c) Displacement

The displacement of the car is the integral of the velocity with respect to time. So, we integrate the equation for velocity from 0 to 6 seconds:

∫v dt = ∫(0.25t^2) dt = (0.25/3)t^3 + C

Again, since the car starts from rest, the constant of integration C = 0. So, the displacement at t = 6 seconds is:

s = (0.25/3)*(6)^3 = 18 m.