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Question

(a) Derive the formula: $$s =ut + \dfrac{1}{2}at^2$$, where the symbols have usual meanings. 
(b) A train starting from stationary position and moving with uniform acceleration attains a speed of $$36 km$$ per hour in $$10$$ minutes. Find its acceleration. 


Solution

(a) Suppose a body has an initial velocity '$$u$$' and a uniform acceleration '$$a$$' for time '$$t$$' so that its final velocity becomes '$$v$$'. Let the distance travelled by the body in this time be '$$s$$'. The distance travelled by a moving body in time '$$t$$' can be found out by considering its average velocity. Since the initial velocity of the body is '$$u$$' and its final velocity is '$$v$$', the average velocity is given by 
Average velocity $$= \dfrac{\text{Initial velocity + Final velocity}}{2}$$
That is, Average velocity $$= \dfrac{u + v}{2}$$ 
Also, Distance travelled $$=$$ Average velocity $$\times$$ Time 
So, $$s = \left(\dfrac{u+v}{2}\right)\times t$$
From the first equation of motion, we have, $$v = u+ at$$.
Put this value of $$v$$ in equation (1), we get: 
$$s =\left( \dfrac{u +u + at}{2}\right) \times t$$
or $$s = \dfrac{(2u +at) \times t}{ 2}$$
or  $$s = \dfrac{2ut +at^2}{2}$$
or $$s =ut +\dfrac{1}{2}at^2$$
where, $$s=$$ distance travelled by the body in time $$t$$
$$u =$$ initial velocity of the body 
and $$a=$$ acceleration 
(b) Initial velocity, $$u=0m/s$$
Final velocity, $$v=36km/h=10m/s$$
Time, $$t=10min =10 \times 60=600\ sec$$
Acceleration $$= \dfrac{\text{Final velocity - Initial velocity}}{\text{time taken}}$$
So, $$a = \dfrac{v-u}{t} = \dfrac{10-0}{600} = \dfrac{10}{600} m/s^2=\dfrac{1}{60}m/s^2 = 0.016 m/s^2$$

Physics

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