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Physics Question on Motion in a plane

A cyclist is riding with a speed of 27 km h−127\,km\, h^{-1}. As he approaches a circular turn on the road of radius 80 m80 \,m, he applies brakes and reduces his speed at the constant rate of 0.50 ms−10.50\,m s^{-1} every second. The net acceleration of the cyclist on the circular turn is

A

0.68 ms−20.68\,ms^{-2}

B

0.86 ms−20.86\,ms^{-2}

C

0.56 ms−20.56\,ms^{-2}

D

0.76 ms−20.76\,ms^{-2}

Answer

0.86 ms−20.86\,ms^{-2}

Explanation

Solution

Here, v=27 km h−1v=27\,km\,h^{-1} =27×518ms−1=27\times\frac{5}{18}ms^{-1} v=152ms−1=7.5 ms−1v=\frac{15}{2}ms^{-1}=7.5\,ms^{-1}, r=80 mr = 80\,m Centripetal acceleration, ac=v2ra_{c}=\frac{v^{2}}{r} ac=(7.5 ms−1)280 m≈0.7 ms−2a_{c}=\frac{\left(7.5\,ms^{-1}\right)^{2}}{80\,m}\approx0.7\,ms^{-2} Tangential acceleration, at=0.5 ms−2a_{t}=0.5\,ms^{-2} Magnitude of the net acceleration is a=(ac)2+(at)2a=\sqrt{\left(a_{c}\right)^{2}+\left(a_{t}\right)^{2}} =(0.7)2+(0.5)2=\sqrt{\left(0.7\right)^{2}+\left(0.5\right)^{2}} ≈0.86 ms−2\approx0.86\,ms^{-2}