Question

A man moves on his motorbike with 54 km/h and then takes a u-turn and continues to move with same speed. The time of u-turn is 10 sec. Find the average acceleration during u-turn.

Answer 1

A man moves on his motorbike with $54\mathrm{km / h}$ and then takes a u-turn and continues to move with same speed. The time of u-turn is 10 sec. Find the average acceleration during u-turn.

Solution:

During the u-turn motorbike is moving in a circle at a constant speed, so it acts centripetal acceleration. In the process of turning a motorbike passes a half of the circle with a radius R:

$$\frac {2 \pi R}{2} = V t; t = 1 0 s, V = 5 4 \frac {k m}{h} = 1 5 \frac {m}{s}$$ $$R = \frac {V t}{\pi} (1)$$

Centripetal acceleration formula:

$$a = a _ {c e n t r} = \frac {V ^ {2}}{R} (2)$$ $$(1) i n (2): a = \frac {\pi V ^ {2}}{V t} = \frac {\pi V}{t} = \frac {3 . 1 4 * 1 5 \frac {m}{s}}{1 0 s} = 4. 7 1 \frac {m}{s ^ {2}}$$

Answer: average acceleration $4.71\frac{m}{s^2}$