Two trains A and B of length \(300m\) each are moving on two parallel tracks with a uniform speed of \(54 kmh^{-1}\) in the same direction, with train a ahead of train B. The loco-pilot of train B decides to overtake the train A and accelerates by \(2 ms^{-1}\). If, after \(25sec\), the guard of train B just brushes past the loco-pilot of train A. What is the original distance between them initially?
Ans: \(25m\).
A man is \(9 m\) behind the door of a train when its starts moving with an acceleration, \(a=2 ms^{-1}\). The man runs at full speed. How far he has to run and after what time he get in to the train?
Ans: \(18m\) and \(3s\).
An object is moving along \((+ve) x-axis\) with a uniform acceleration of \(4 ms^{-2}\). At time, \(t=0 Sec\), \(x=5 m\) and \(v=3 ms^{-1}\). (a) What will be the velocity and position of the object at time, \(t=2 Sec\)? (b) What will be the position of the object when it is has a velocity of \(5 ms^{-1}\)?
Ans:(a) \(v_{t=2s}=11m.s^{-1}\) and \(x_{t=2s}=19m\). (b) \(x_{v=5m.s^{-1}}=7m\).
The displacement of a particle moving in one dimension, under the action of constant force is related to time, \(t\) by equation \(t=\sqrt{x}+3\). Where; x is in meter and t is in seconds. Find the displacement of the particle when it’s velocity is \(Zero\).
Ans: \(displacement=0\left(Zero\right)\).
The velocity of a particle is given by, \(v=v_{0}+gt+ft^{2}\). If, its position is \(x=0\) at \(t=0\), then what is the displacement of the particle after \(t=1s\)?
Ans: \(x=v_{o}+\frac{1}{2}g+\frac{1}{3}f\).
A bus is moving with a speed of \(10 ms^{-1}\) on a straight road. A scooter wishes to overtake the bus in \(100s\). If, the bus is at a distance of \(1 km\)from the scooterist, with what speed could it chase the bus?
Ans: \(v_{s}=20m.s^{-1}\).
Two boys are standing at the ends A and B of a ground where, \(AB=a\). The boy at B starts running in a direction perpendicular to the AB with velocity \(v_{1}\). The boy at A starts running simultaneously with a velocity \(v\) and catches the other boy in time \(t\). Find the value of t?
Ans: \(t=\frac{a}{\sqrt{v^{2}-v_{1}^{2}}}\).
A particle moving along x-axis has acceleration \(f\) at time \(t\) is given by, \(f=f_{0}\left(1-\frac{t}{T}\right)\). Where; \(f_{0}\) and \(T\) are constants. The particle at \(t=0\) has \(Zero\) velocity. Find the velocity of the particle in the time interval between \(t=0\) and the instant when \(f=0\).
Ans: \(v=\frac{1}{2}f_{o}T\).
A particle is moving along x-axis. The position of the particle at any instant of the time is given by, \(x=a+bt^{2}\). Where; \(a=6m\) and \(b=3.5 ms^{-1}\). \(t\) is measured in seconds. Find: (i) Velocity of the particle at \(t=0s\) and \(t=3s\). (ii) Avg. velocity between \(t=3s\) and \(t=6s\).
Ans:(i) \(v_{t=0s}=0m.s^{-1}\) and \(v_{t=3s}=21m.s^{-1}\). (ii) \(v_{avg}=31.5m.s^{-1}\).
