Question

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

• A. $0.5m/s^2$
• B. $0.87m/s^2$
• C. $0.56m/s^2$
• D. $1m/s^2$

Answer 1

The correct option is B $0.87m/s^2$

Given,

$v=27km/h=\frac{27\times 1000}{60\times 60}=7.5m/s$

$r=80m$

$a_t=0.50m/s^2$

Radial acceleration , $a_r=\frac{v^2}{r}$

$a_r=\frac{7.5\times 7.5}{80}=0.7031m/s^2$

The net acceleration of cyclist on the circular turn is,

$a=\sqrt{a_r^2+a_t^2}$

$a=\sqrt{(0.7031{)}^2+(0.50{)}^2}=\sqrt{0.744}m/s^2$

$a=0.87m/s^2$

The correct option is B.