Question

A cyclist is riding with a speed of 27 km $h^{- 1}$ . As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 ${ms}^{- 1}$ every second. The net acceleration of the cyclist on the circular turn is

Moderate

Solution

Here, v = 27 km $h^{- 1}$ = 27 $\times$ $\frac{5}{18} {ms}^{- 1}$
v = $\frac{15}{2} = 7.5 {ms}^{- 1}$
r = 80 m
Centripetal acceleration, $a_{c} = \frac{v^{2}}{r}$
$a_{c} = \frac{{(7.5 {ms}^{- 1})}^{2}}{80 m} = 0.7 {ms}^{- 2}$
Tangential acceleration, at = 0.5 ${ms}^{- 2}$
Magnitude of the net acceleration is
a = $\sqrt{{(a_{c})}^{2} + {(a_{t})}^{2}} = \sqrt{{(0.7)}^{2} + {(0.5)}^{2}} = 0.86 {ms}^{- 2}$

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