Question

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

• A. $0.68m{s}^{-2}$
• B. $0.86m{s}^{-2}$
• C. $0.56m{s}^{-2}$
• D. $0.76m{s}^{-2}$

Answer 1

Answer: (B)

Here, $v=27km{h}^{-1}$
$=27\times \frac{5}{18}m{s}^{-1}$
$v=\frac{15}{2}m{s}^{-1}=7.5m{s}^{-1}$,
$r=80m$
Centripetal acceleration, $a_c=\frac{v^2}{r}$
$a_c=\frac{{\left(7.5m{s}^{-1}\right)}^2}{80m}\approx 0.7m{s}^{-2}$
Tangential acceleration, $a_t=0.5m{s}^{-2}$
Magnitude of the net acceleration is
$a=\sqrt{{\left(a_c\right)}^2+{\left(a_t\right)}^2}$
$=\sqrt{{\left(0.7\right)}^2+{\left(0.5\right)}^2}$
$\approx 0.86m{s}^{-2}$