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 0.05 m s^-1 every second. The net acceleration of the cyclist on the circular turn is

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

Answer 1

Answer: (B)

$0.86ms^{- 2}$ ($0.86m/s^{2}$)

Answer 2

Here, v = 27 km $h^{- 1} = 27 \times \frac{5}{18}ms^{- 1}$

$v = \frac{15}{2}ms^{- 1} = 7.5ms^{- 1}$

R = 80m

Centripetal accelerations, $a_{x} = \frac{v^{2}}{r}$

$a_{c} = \frac{(7.5ms^{- 1})^{2}}{80m} = 0.7ms^{- 2}$

Tangential acceleration, $a_{1} = 0.5ms^{- 2}$

Magnitude of the net accelerations is

$a = \sqrt{(a_{c})^{2} + (a_{t})^{2}} = \sqrt{(0.7)^{2} + (0.5)^{2}} = 0.86ms^{- 2}$