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Question

A particle moving with a non-zero velocity and constant acceleration on straight line travels $$15\ m$$ in first $$3\ s$$ and $$33\ m$$ in next $$3\ s$$ in the same direction, find
(i) initial velocity of particle
(ii) acceleration of particle


Solution

Let initial velocity $$u$$ and acceleration $$a$$. 
Using $$s=ut +\dfrac{1}{2}at^2$$
$$\Rightarrow 15= 3u + \dfrac{1}{2}\times a \times 3^2$$    $$\Rightarrow 6u+9a= 30$$      .....1
 Distance travelled in $$6s$$, $$=15+33=48 m$$
$$\Rightarrow 48= 6u +\dfrac{1}{2}\times a\times 6^2$$   $$\Rightarrow 6u + 18a= 48$$     .....2
Subtracting equation 1 from 2, we get  $$\Rightarrow 9a=18$$   $$\Rightarrow a=2 m/s^2$$
Put the value of $$a=2$$ in equation 1, we get $$\Rightarrow u=2m/s$$
Hence, Initial velocity $$= 2 m/s$$
Acceleration $$= 2 m/s^2$$

Physics

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