A person standing on the floor of an elevator drops a coin. The coin reaches the floor in time $t_1$ if the elevator is moving uniformly and time $t_2$ if the elevator is stationary. Then:

A person travelling in a straight line moves with a constant velocity ${\mathrm{v}}_1$ for a certain distance 'x' and with a constant velocity ${\mathrm{v}}_2$ for the next equal distance. The average velocity v is given by the relation:

A particle moves along a path ABCD as shown in the figure. The magnitude of the displacement of the particle from A to D is:

The velocity of a particle moving along a straight line with constant acceleration 'a' reduces to $\frac{1}{5}$ of its initial velocity in time $\text{'}\tau \text{'}$. The total time taken by the body till its velocity becomes zero is-

A boy falls from a building of height 320 m. After 5 seconds, superman jumps downward with initial speed u such that the boy can be saved. The minimum value of u is (assume g=10 $\mathrm{m}/{\mathrm{s}}^2$)

The displacement x of a body having relation with time as $\mathrm{x}=3{\mathrm{t}}^2-6\mathrm{t}+7$, then distance covered by body in first two second is [where x is in metre and t is in second].

The acceleration of particle for which the ${\mathrm{v}}^2$ versus s graph is shown is:

In the following questions, a statement of assertion (A) is followed by a statement of the reason (R).

A: A body is momentarily at rest when it reverses the direction of motion.

R: A body cannot have acceleration if its velocity is zero at a given instant of time.

The position (x) of a particle moving along x-axis varies with time (t) as $\mathrm{x}={\mathrm{t}}^2-4\mathrm{t}+5$. The particle turns around at

The velocity (v) of a particle varies with position (x) as shown in the graph. Its acceleration when x=4 m will be

The acceleration time graph of a particle moving along a straight line as shown in figure. The time taken by the particle to acquire its initial velocity is?

Position of a particle moving on a straight line is given by $\mathrm{x}=2{\mathrm{t}}^2-6\mathrm{t}$. The distance travelled by the particle in first four seconds is

A bird flies for 4 s with a velocity of |t-2| m/s in a straight line, where t is time in seconds. It covers a distance of

The acceleration (a) versus time (t) graph of a particle moving along a straight line is as shown in the given figure. The particle starts from rest at t=0. The velocity of the particle is maximum at

A graph between square of speed with time t for a particle moving on a straight line. The acceleration of the particle at t=1 s is

A body moved in a straight line according to equation $\mathrm{x}=30\mathrm{t}-5{\mathrm{t}}^2$. Th value of $\frac{\left|\mathrm{displacement}\right|}{\mathrm{distance}}$ between t=2 s to t=5 s is

A particle is moving with a velocity of $\mathrm{v}=(3+6\mathrm{t}) {\mathrm{cms}}^{-1}$. The displacement of the particle in the interval t=0 sec to t=5 sec is:

A ball is thrown vertically downwards with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with the velocity of 80 m/s . The height of the tower is: $(assuming \mathrm{g}=10 \mathrm{m}/{\mathrm{s}}^2)$

A person sitting on the ground floor of a building notices through the window, of height 1.5 m, a ball dropped from the roof of the building crosses the window in 0.1 s. What is the velocity of the ball when it is at the topmost point of the window? (g = 10 m/${\mathrm{s}}^2$)

Which one is not an example of a point object?

A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backwards, followed again by 5 steps forward and 3 steps backwards, and so on. Each step is 1 m long and requires 1 s. There is a pit on the road 13 m away from the starting point. The drunkard will fall into the pit after:

The speed-time graph of a particle moving along a fixed direction is shown in the figure. Then the distance traversed by the particle between t = 0 sec to 10 sec:

$$\begin{gathered} \left(1\right) 70 m \\ \left(2\right) 60 m \\ \left(3\right) 50 m \\ \left(4\right) 40 m \end{gathered}$$

The velocity-time graph of a particle in one-dimensional motion is shown in Figure. Which of the following formulae is correct for describing the motion of the particle over the time interval $t_1$ to $t_2$?

$$\begin{gathered} \left(1\right) \mathrm{x}({\mathrm{t}}_2 ) = \\ \mathrm{x}\left({\mathrm{t}}_1\right) + \mathrm{v}\left({\mathrm{t}}_1\right)\times ({\mathrm{t}}_2–{\mathrm{t}}_1) +\frac{1}{2} \mathrm{a} {({\mathrm{t}}_2–{\mathrm{t}}_1)}^2 \\ \left(2\right) \mathrm{v}({\mathrm{t}}_2 ) = \\ \mathrm{v}\left({\mathrm{t}}_1\right) + \mathrm{a} ({\mathrm{t}}_2–{\mathrm{t}}_1) \\ \left(3\right) \mathrm{x}({\mathrm{t}}_2 ) = \\ \mathrm{x}\left({\mathrm{t}}_1\right) + {\mathrm{v}}_{\mathrm{average}}\times ({\mathrm{t}}_2–{\mathrm{t}}_1) +\frac{1}{2} {\mathrm{a}}_{\mathrm{average}}\times {({\mathrm{t}}_2–{\mathrm{t}}_1)}^2 \\ \left(4\right) {\mathrm{v}}_{\mathrm{average}} = \\ \frac{ \left(\mathrm{x}\right({\mathrm{t}}_2) – \mathrm{x}({\mathrm{t}}_1\left)\right)}{\left({\mathrm{t}}_2-{\mathrm{t}}_1\right)} \end{gathered}$$

The figure gives the x-t plot of a particle executing a one-dimensional simple harmonic motion. Then the signs of position & velocity variables of the particle at t = -1.2 sec respectively are:

$$\begin{gathered} \left(1\right) \mathrm{positive}, \mathrm{negative} \\ \left(2\right) \mathrm{positive}, \mathrm{positive} \\ \left(3\right) \mathrm{negative}, \mathrm{positive} \\ \left(4\right) \mathrm{negative}, \mathrm{negative} \end{gathered}$$

The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in the graph. Choose the incorrect statement.

A jet airplane travelling at the speed of $500 km h^{-1}$ ejects its products of combustion at the speed of $1500 km h^{-1}$ relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?

$$\begin{gathered} 1. 1000 km h^{-1} \\ 2. 500 km h^{-1} \\ 3. 1500 km h^{-1} \\ 4. 2000 km h^{-1} \end{gathered}$$

A car moving along a straight highway with a speed of $126 km h^{-1}$ is brought to a stop within a distance of 200 m. How long does it take for the car to stop?

Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of $72 km h^{-1}$ in the same direction with A ahead of B. The driver of B decides to overtake A and accelerates by $1 m s^{-2}$. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?

On a two-lane road, car A is travelling at a speed of $36 km h^{-1}$. Two cars B and C approach car A in opposite directions with a speed of $54 km h^{-1}$ each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?

$$\begin{gathered} \left(1\right) 1 m s^{-2} \\ \left(2\right) 5 m s^{-2} \\ \left(3\right) 2 m s^{-2} \\ \left(4\right) 3 m s^{-2} \end{gathered}$$