Question

A cyclist is moving with a speed of $6m{s}^{-1}$ . As he approaches a circular turn on the road of radius $120m$ , he applies brakes and reduces his speed at a constant rate of $0.4m{s}^{-2}$ . The magnitude of the net acceleration of the cyclist on the circular turn just after brakes is

• A. $0.5m{s}^{-2}$
• B. $1.0m{s}^{-2}$
• C. $2.0m{s}^{-2}$
• D. $4.0m{s}^{-2}$

Answer 1

Answer: (A)

It is a type of non-uniform circular case, so it has both radial $\left(a_c\right)$ and tangential acceleration $\left(a_t\right)$ , and net acceleration $\left(a_{net}\right)$ is the vector sum of both acceleration,
$a_t=0.4m{s}^2$
$a_c=\frac{F}{m}$
$a_c=\frac{m{v}^2}{mr}$
$a_c=\frac{v^2}{r}\Rightarrow \frac{6^2}{120}\Rightarrow 0.3m{s}^{-2}$
$a_{net}=\sqrt{a_c^2+a_t^2}=\sqrt{(0.3{)}^2+(0.4{)}^2}=0.5$