List of top Physics Questions on Motion in a straight line asked in AP EAPCET

A particle moves with velocity $v = at - bt^2$ where a and b are constants. The acceleration becomes zero at:

A particle under constant acceleration covers the first half of the total distance in time $t_1$ and remaining in time $t_2$. Then:

A particle moves along a straight line along the \(x\)-axis. Its position \((x)\) versus time \((t)\) graph is shown in the figure \([x \text{ in meters and } t \text{ in seconds}]\). Its average speed during the motion is:

A student is at a distance \(16\,\text{m}\) from a bus when the bus begins to move with a constant acceleration of \(9\,\text{m s}^{-2}\). The minimum velocity with which the student should run towards the bus so as to catch it is \(\alpha\sqrt{2}\,\text{m s}^{-1}\). The value of \(\alpha\) is

\(y=(Pt^2-Qt^3)\;m\) is the vertical displacement of a ball which is moving in vertical plane. Then the maximum height that the ball can reach is

An object moving along \(x\)-axis with a uniform acceleration has velocity \[ \vec{V}=(12\;\text{cm s}^{-1})\hat{i} \] at \(x=3\;\text{cm}\). After \(2\;\text{s}\), if it is at \(x=-5\;\text{cm}\), then its acceleration is

A passenger sees a bus standing 10 m ahead of him. The bus starts moving with a constant acceleration of \(5 \, m\,s^{-2}\) away from him. The passenger runs with constant speed to catch the bus. The minimum speed required to catch the bus is:

A particle is moving along straight line. If its location at time t is given by \(x(t) = \alpha t e^{-t/\tau}; \alpha = 1~\text{m/s}, \tau = 1~\text{s}\). The velocity of the particle at t = 2 s is

A ball is dropped from a tower of height \(80\,\text{m}\). The time it takes to cover the last \(50\%\) of its fall is (acceleration due to gravity \(=10\,\text{m s}^{-2}\))