Question

1.A biker is riding his bike at the speed of 54 km/h. As the rider approaches the circular turn on the road having radius 70 m, he reduces the speed of bike at a constant rate of 0.70 m/s2, by applying brakes. Find out the direction and magnitude of the net acceleration of the biker on this turn.

Answer 1

Given:

A biker is riding his bike at a speed of 54 km/hr. As the Ryder approaches a circular turn of radius 70 metres, he reduces the speed at a constant rate of 0.7 m/s².

To find:

Net magnitude and direction of acceleration.

Calculation:

Net accelerationwill be vector addition of the centripetal acceleration and the tangential acceleration experienced by the bike in the circular trajectory.

54 km/hr = 54 × (5/18) = 15 m/s

Tangential acceleration = 0.7 m/s².

Centripetal acceleration = v²/R

=> Centripetal acceleration = (15)²/70

=> Centripetal acceleration = 3.21 m/s².

Net acceleration be a:

$a = \sqrt{ {(3.21)}^{2} + {(0.7)}^{2} }$

$= > a = \sqrt{ 10.3 + 0.49 }$

$= > a = \sqrt{ 10.79 }$

$\boxed{ = > a = 3.28 \: m {s}^{ - 2} }$

Let angle between resultant acceleration and Centripetal acceleration be $\theta$.

$\therefore \: \tan( \theta) = \dfrac{0.7}{3.21}$

$= > \: \tan( \theta) = 0.22$

$\boxed{= > \: \theta = 12.4 \degree}$