Question

A scooterist is approaching a circular turn of radius 80 m. He reduced his speed from 27 kmh^-1 at constant rate of 0.5 ms^-2. His vector acceleration on the circular turn is

• A. 0.86 ms^-2,54°
• B. 0.68 ms^-2,45⁰
• C. 1.0 ms^-2,45⁰
• D. 0.5 ms^-2,45⁰

Answer 1

Answer: (A)

0.86 ms^-2,54°

Answer 2

Centripetal acceleration

a_r = $\frac{v^{2}}{r} = \frac{\left( \frac{27 \times 1000}{3600} \right)^{2}}{80}$

= $\frac{7.5 \times 7.5}{80}$ = 0.703 ms^-2

Net acceleration, a = $\sqrt{a_{r}^{2} + a_{t}^{2}}$

where a_t = tangential acceleration = 0.5 ms^-2

∴ a = $\sqrt{0.703^{2} + 0.5^{2}} = 0.86ms^{- 2}$and

θ = tan^-1 $\frac{a_{r}}{a_{t}} = \tan^{- 1}\frac{0.703}{0.5} = 54^{o}$