The displacement of a particle starting from rest \(\left(t=0\right)\) is given by, \(x=3t^{2}-t^{3}\). Calculate: the time at which acceleration of the particle becomes Zero.
Ans: \(t=1Sec\).
The velocity of a particle is given by the equation, \(v=\left(2t^{2}+5\right) cms^{1}\). Find: (i)The change in velocity of the particle during the time interval between, \(t_{1}=2s\) and \(t_{2}=4s\). (ii) The average acceleration during same time interval. (iii) The instantaneous acceleration at \(t_{2}=4s\).
Ans:(i) \(v_{2}-v_{1}=24 cm.s^{-1}\). (ii) \(a_{avg}=12 cm.s^{-2}\) and (iii) \(a_{t=4sec}=16 cm.s^{-2}\).
A particle is moving along X-axis. The position of a particle at any instant of time is given by, \(x=10+0.0t^{2}\). Where; x is measured in meters and t is in seconds. Find:(i) Average acceleration of the particle between \(t=2sec\) and \(t=3s\). (ii) Also, show that acceleration of the particle is constant.
Ans:(i) \(a_{avg}=0.4m.s^{-2}\) and (ii) \(a_{ins}=0.4m.s^{-2}\left(Constant\right)\).
A car starts from rest and acquires a velocity of \(54 kmh^{-1}\) after \(20Sec\). What will be its acceleration of the car?
Ans: \(a=\frac{3}{4}m.s^{-1}\).
An electron is emitted with a velocity of \(5\times10^{6} ms^{-1}\). It is accelerated by an electric field in the direction of initial velocity at \(3\times10^{14} ms^{-2}\). If, its final velocity is \(7\times10^{6} ms^{-1}\) . Calculate: the time taken by the electron to attain the final velocity and distance covered by it.
Ans: \(t=\left(6.67\times 10^{-9}\right)Sec\) and (ii) \(S=0.04m\).
A boy starts moving from rest and acquires a velocity of \(12 ms^{-1}\) in \(5Sec\). Calculate:the acceleration and distance covered by the body?
Ans: \(a=2.4m.s^{-2}\) and \(S=30m\).
A body is moving with uniform acceleration covers \(20m\) in \(2^{nd} Sec\) and \(30m\) in \(4^{th} Sec\) of the motion. Calculate distance covered in \(6^{th}Sec\).
Ans: \(S_{6^{th}}=40m\).
Brakes are applied to a car travelling at \(30 ms^{-1}\). Its velocity is reduced to \(20ms^{-1}\) in \(5Sec\). Calculate: (i)Retardation produced by the brakes. (ii) Distance covered is \(5Sec\).
Ans:(i) \(a=-2m.s^{-2}\) and (ii) \(S=125m\).
A ball is thrown vertically upwards with a velocity of \(20 ms^{-1}\) from the top of a tower. The height of the point from where the ball is thrown is \(25 m\) from the ground. Then; (a) How height the ball will rise? and (b) How long will be before the ball hits the ground. Take: g=\(10 ms^{-2}\).
