Question

# A body which is uniformly accelerated, changes its velocity from $$36 km/hr$$ in one direction to $$18 km/hr$$ in the opposite in $$6$$ seconds. The total distance traveled by the body during this time interval is

Solution

## Initial velocity, $$\begin{array}{l}u = 36 \times \frac{5}{{18}}\\ = 10\,m{s^1}\end{array}$$velocity after $$\begin{array}{l}6s = 18 \times \frac{5}{{18}}\\ = 5\,m{s^1}\end{array}$$$$\begin{array}{l}v = u + at\\ - 5 = 10 + a \times 6\\a = - 2.5\,m{s^1}\end{array}$$Let distance travelled to the right side $${s_1}$$$$\begin{array}{l}{v^2} = {u^2} + 2a{s_1}\\0 = {10^2} + 2\left( {-2.5} \right){s_1}\\{s_1} = 20\,m\end{array}$$Let distance travelled to the left side be $${s_2}$$  $$\begin{array}{l}{5^2} = 2 \times 2.5 \times {s_2}\\{s_2} = 5m\end{array}$$Total distance $$\begin{array}{l} = 20 + 5\\ = 25m\end{array}$$Physics

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