Question

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

• A. $0.5m/s^2$
• B. $0.86m/s^2$
• C. $0.56m/s^2$
• D. $1m/s^2$

Answer 1

The correct option is B $0.86m/s^2$

$v=27Km/hr$

$=\frac{27\times 1000}{3600}=7.5m/sec$

Centripetal acceleration = $\frac{(7.5{)}^2}{80}=0.7m/s^2$

Tangential acceleration = $-0.5m/s^2$

$\therefore a_{net}=\sqrt{(0.7{)}^2+(0.5{)}^2}=0.86m/s^2$