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 $0.50\,m s^{-1}$ every second. The net acceleration of the cyclist on the circular turn is

• A. $0.68\,ms^{-2}$
• B. $0.86\,ms^{-2}$
• C. $0.56\,ms^{-2}$
• D. $0.76\,ms^{-2}$

Answer 1

The Correct Option is B

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