Question

A jeep runs around the curve of radius 0.3 km at a constant speed of 60 ms^-1. The resultant change in velocity, instantaneous acceleration and average acceleration over 60⁰ arc are

• A. 30 ms^-1, 11.5 ms^-2,12 ms^-2
• B. 60ms^-1, 12 ms^-2,11.5ms^-2
• C. 60ms^-1, 11.5 ms^-2,12ms^-2
• D. 40 ms^-1, 10ms^-2,8ms^-2

Answer 1

Answer: (B)

60ms^-1, 12 ms^-2,11.5ms^-2

Answer 2

∆$\overrightarrow{v} = 2v\sin\left( \frac{\theta}{2} \right)$ = 2 × 60 sin 30 = 60 ms^-1

Instantaneous acceleration

=$\frac{v^{2}}{r} = \frac{60^{2}}{0.3 \times 1000} = 12ms^{- 2}$

Time taken to cover the arc

= t = $\frac{\pi}{3} \times \frac{300}{60}$

using ∆t = $\frac{S}{v} = \frac{rd\theta}{v}$

∴ average acceleration a = $\frac{\Delta v}{t}$ = $\frac{60}{\frac{\pi}{3} \times \frac{300}{60}} = 11.5ms^{- 2}$