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Question

(a) Derive the formula: s=ut+12at2, where the symbols have usual meanings.
(b) A train starting from stationary position and moving with uniform acceleration attains a speed of 36km per hour in 10 minutes. Find its acceleration.

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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 =Initial velocity + Final velocity2
That is, Average velocity =u+v2
Also, Distance travelled = Average velocity × Time
So, s=(u+v2)×t
From the first equation of motion, we have, v=u+at.
Put this value of v in equation (1), we get:
s=(u+u+at2)×t
or s=(2u+at)×t2
or s=2ut+at22
or s=ut+12at2
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=10×60=600 sec
Acceleration =Final velocity - Initial velocitytime taken
So, a=vut=100600=10600m/s2=160m/s2=0.016m/s2

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