Test: Uniform Circular Motion (NCERT) - NEET MCQ

The term centripetal acceleration was proposed by Newton.

Centripetal acceleration, a_c = v²/R
where v is the speed of an object and R is the radius of the circle. It is always directed towards the center of the circle. Since v and R are constants for a given uniform circular motion, therefore the magnitude of centripetal acceleration is also constant. However, the direction of centripetal acceleration changes continuously. Therefore, centripetal acceleration is not a constant vector.

Here, v  = 400ms⁻¹,

r = 0.6 km = 0.6 × 10³m,

g = 10ms⁻²
Centripetal acceleration,

= 26.7 × 10ms⁻²
In the units of (g = 10ms⁻²), the centripetal acceleration is 26.7.

In a uniform circular motion, the acceleration is directed towards the centre while velocity is acting tangentially. Therefore, velocity vector is perpendicular to acceleration vector.

In Circular motion, the Position vector is always outward and passes through the center. whereas the and centripetal acceleration is always toward the center. 
∴ Both is the opposite direction.

For a particle performing uniform circular motion, magnitude of the acceleration remains constant.

The kinematic equations for uniform acceleration do not apply in case of uniform circular motion because in this case the magnitude of acceleration is constant but its direction is changing.

Let the particle is moving from point A to B
Displacement of the particle = AB
According to cosine formula

Displacement AB = x = r

Since the initial position coincides with the final position. So, the net displacement of the cyclist = zero

Given here,
Speed of the cyclist, v = 27 km/h = 7.5m/s
Radius of the circular tun on the road, r = 80 m
Therefore the centripetal acceleration of the cyclist at the moment when velocity is 7.5 m/s is,

When brakes are applied, the speed is decreased at the rate of 0.5m/s every second.

Therefore the tangential acceleration is,
a_r = - 0.5 m/s²
Therefore the total acceleration of the cyclist is,