Question

A cyclist is riding with a speed of $ 27km/h $ . As he approaches a circular turn on the road of radius $ 80m $ , he applies brakes and reduces his speed at the constant rate of $ 0.5m/s $ every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?

Answer 1

Hint :For solving this question, you need to recall the standard formulae of translational motion and rotational motion. The time of motion isn’t provided in the question, so we need to use a formula that only uses velocity and the radius to find out acceleration.

Complete Step By Step Answer:
Net acceleration is due to braking and centripetal acceleration. The net acceleration is the vector sum of braking and centripetal acceleration, and the direction angle is the inverse tangent value of the centripetal acceleration and the tangential acceleration.
Due to Braking, Tangential Acceleration: $ {a_t} = 0.5m/{s^2} $
Speed of the cyclist: $ v = 27km/h $
Now, $ 27 \times \dfrac{5}{{18}} = 7.5m/s $
Radius of the circular turn: $ r = 80m $
Centripetal acceleration is given as:
  $ {a_c} = \dfrac{{{v^2}}}{r} = \dfrac{{{{\left( {7.5} \right)}^2}}}{{80}} \\
  {a_c} = \dfrac{{56.25}}{{80}} \approx 0.70m/{s^2} \\ $
Since the angle between $ {a_c} $ ​ and $ {a_t} $ ​ is $ {90^ \circ } $ , the resultant acceleration $ a $ is given by:
  $ a = \sqrt {{a_c}^2 + {a_t}^2} \\
  a = \sqrt {{{0.7}^2} + {{0.5}^2}} = 0.86m/{s^2} \\ $
Now, $ \tan \theta = \dfrac{{{a_c}}}{{{a_t}}} $ , where $ \theta $ is the angle of the resultant with the direction of the velocity.
  $ \tan \theta = \dfrac{{0.7}}{{0.5}} = 1.4 \\
  \theta = {\tan ^{ - 1}}\left( {1.4} \right) = {54.56^ \circ } \\ $

Note :
It is important to understand that the deceleration mentioned in the question acts only on the tangential acceleration, which is a common mistake made by students. Since tangential acceleration is not assigned in the question, we take the deceleration as the tangential acceleration. Also, the angle we found out is the angle between the tangential acceleration and the net acceleration.