Question

A cyclist is riding with a speed of 18 kmh-¹. As he approaches a circular turn on the road of radius 25√2 m, he applies brakes and reduces his speed at the constant rate of 0.5 ms-¹ every second. Determine the magnitude and direction
of the net acceleration of the cyclist on the circular turn.
​

Answer 1

Solution :

Given :

A cyclist is riding with a speed of 18 kmh-¹. As he approaches a circular turn on the road of radius 25√2 m, he applies brakes and reduces his speed at the constant rate of 0.5 ms-¹ every second.

To find :

The magnitude and direction

of the net acceleration of the cyclist on the circular turn.

Answer :

Velocity = 18km/h

Velocity = $\sf{18\:\times\:\dfrac{5}{18}\:=\:5m/s^{-1}}$

We have Acceleration towards the centre , $\sf{\dfrac{v^2}{r}}$

Acceleration tendency = $\sf{a_t\:=\:\dfrac{d_v}{d_t}\:=\:\dfrac{1}{2}ms^{-2}}$

Centripetal acceleration = $\sf{\dfrac{v^2}{r}\:=\:\dfrac{25}{25\sqrt{2}}\:=\:\dfrac{1}{\sqrt{2}}m/s^{-2}}$

$\sf{a_{net}\:=\:\sqrt{a_t^2\:+\:a_c^2}}$

= $\sf{\sqrt{\dfrac{1}{2}\:+\:\dfrac{1}{4}}}$

= $\sf{\dfrac{\sqrt{3}}{2}\:m/s^{-2}}$

$\sf{\tan\theta\:=\:\dfrac{a_c}{a_t}\:=\:\dfrac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\:=\:\dfrac{2}{\sqrt{2}}\:=\:\sqrt{2}}$

$\sf{\theta\:=\:\tan^{-1}\:(\sqrt{2})}$

Answer 2

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Speed of the cyclist ,

v= 18km/hr

$\implies \: v = 18 \times \frac{5}{18} m {s}^{ - 1}$

As the cyclist approaches the turn..

The tangential acceleration

$at = 0.5m {s}^{ - 2}$

And the radial acceleration:-

$ac = \frac{ {v}^{2} }{r} = \frac{25}{52 \sqrt{2} } \\ \implies \: ac = \frac{1}{ \sqrt{2} } m {s}^{ - 2}$

Thus..

Total acceleration

$aT = \sqrt{ {a}^{2}c + {a}^{2}t } \\ \implies \: aT = \sqrt{( { \frac{1}{ \sqrt{2} }) }^{2} + { \frac{1}{2} }^{2} } \\ \implies \: aT = \sqrt{ \frac{1}{2} + \frac{1}{4} } \\ \implies \: aT = \sqrt{ \frac{3}{4} } \\ \implies \: aT = \frac{ \sqrt{3} }{2} m {s}^{ - 2}$

And the Direction..

$\tan \theta = \frac{ac}{at} \\ \implies \: \tan\theta \: = \frac{ \frac{1}{ \sqrt{2} } }{0.5} \\ \implies \: \tan \theta = \frac{1}{ \sqrt{2} } \times \frac{2}{1} \\ \implies \tan \theta = \sqrt{2} \\ \\ so.. \theta = { \tan }^{ - 1} \sqrt{2}$

hence,

magnitude of net acceleration of the cyclist is

$\bold{= \frac{1 }{ \sqrt{2} } m {s}^{ - 2} }$

and direction of net acceleration of cyclist is

$\bold{= { \tan }^{ - 1} \sqrt{2}}$