The relation between the displacement ' $x$ ' (in metre) and the time ' $t$ ' (in second) of a particle is $t=2 x^2+3 x$. If the displacement of the particle is 25 cm from the origin $(x=0)$, then the acceleration of the particle is

A person wearing a parachute jumps off a plane from a height of 2 km from the ground and falls freely for 20 m before his parachute opens. After his parachute opens if he continues to move uniformly with the velocity attained due to his freefall, the total time taken by the person to reach the ground is (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

A particle initially at rest is moving along a straight line with an acceleration of $2 \mathrm{~ms}^{-2}$. At a time of 3 s after the beginning of motion, the direction of acceleration is reversed. The time from the beginning of the motion in which the particle returns to its initial position is

The ratio of times taken by a freely falling body to travel first 5 m , second 5 m , third 5 m distances is

The driver of a bus moving with a velocity of $72 \mathrm{~km} / \mathrm{h}$ observes a boy walking across the road at a distance of 50 m in front of the bus and decelerates the bus at $5 \mathrm{~ms}^{-2}$ by applying brakes and is just able to avoid an accident. The reaction time of the driver is

The initial and final velocities of a body projected vertically from the ground are $20 \mathrm{~ms}^{-1}$ and $18 \mathrm{~ms}^{-1}$ respectively. The maximum height reached by the body is (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

The displacement-time graphs of two moving particles make angles of $30^{\circ}$ and $45^{\circ}$ with the time axis. The ratio of their velocities is

The acceleration of a vertically projected body at its highest reaching position is

A bird flies with a velocity $(t-2) \mathrm{ms}^{-1}$ along a straight line, where $t$ is the time in seconds. The distance covered by it in a time of 4 seconds is

A body starts rest with uniform acceleration. If its velocity after $n^{\text {th }}$ second (last second) is $v$ then its displacement in the last two seconds is

The ratio of the displacements of a freely falling body during first, second and third seconds of its motion is

A car starts at time $t=0$ from an initial speed of $10 \mathrm{~m} / \mathrm{s}$ and accelerates uniformly with $2 \mathrm{~m} / \mathrm{s}^2$ on a straight road for time $0 \leq t \leq 10 \mathrm{~s}$. Let $s_1$ and $s_2$ be the distance covered by the car in time $3 \leq t \leq 4 \mathrm{~s}$ and $4 \leq t \leq 5 \mathrm{~s}$, respectively. The ratio $\frac{s_2}{s_1}$ is

A ball projected up passes the same height $H$ at 2 s and 10 s . The value of $H$ is (use, $g=9.8 \mathrm{~m} / \mathrm{s}^2$ )

Two towns $X$ and $Y$ are connected by a regular bus service. A bus leaves in either direction at every $t=T$ minutes. A man moving with same speed in the direction $X$ to $Y$ find that a bus goes past him every $t=t_1$ minutes in the direction of his motion, and every $t=t_2$ minutes in the opposite direction. Then, $T$ is given by

A rocket moves straight upward with zero initial velocity and with an acceleration $20 \mathrm{~m} / \mathrm{s}^2$. It runs out of fuel and stops accelerating at the end of 5th second. It reaches a maximum height and falls back to the earth. The speed when it hits the ground is (take $g=10 \mathrm{~m} / \mathrm{s}^2$ )

A particle moves along a straight line, such that its displacement $x$ varies with time $t$ as $x=\alpha t^3+\beta t^2+\gamma$, where $\alpha, \beta$ and $\gamma$ are constants, $v_1$ is the average velocity of the particle during its journey between $t=1 \mathrm{~s}$ and $t=3 \mathrm{~s} . v_2$ is the instantaneous velocity of the particle at $t=3 \mathrm{~s}$. The ratio $\frac{v_1}{v_2}$ is

An aircraft is flying at a height of $h$ above the ground and at a speed of $v$. The maximum angle subtended at a ground observation point by the aircraft after time $t$ is

A particle starts from rest. Its acceleration (a) versus time $(t)$ graph is as shown in the figure. The maximum speed of the particle will be

Assertion (A) The zero velocity of a particle at any instant always implies zero acceleration at that instant.

Reason (R) A body is momentarily at rest when it reverses its direction of motion. The correct option among the following is

A river has a steady speed of $v$. A man swims upstream at a distance of $d$ and swims back to the starting point in total time $t$. The man can swim at a speed of $2 v$ in still water. If the time taken by the man in still water is $t_0$ to complete the same length of swim, then $\frac{t}{t_0}$ is

A body starts from the rest and acquires a velocity of $10 \mathrm{~m} / \mathrm{s}$ in 2s. What is the acceleration of the body and the distance travelled?

A bullet fired into a target losses one-third of its velocity after travelling a distance $x$ metre into the target. If the bullet comes to rest by travelling a further distance $x^{\prime}$, then the ratio $\frac{x^{\prime}}{x}$ is

Consider that a truck is moving initially with $54 \mathrm{~km} / \mathrm{h}$. It has stopped by the driver after looking at an obstacle with a deceleration of $10 \mathrm{~m} / \mathrm{s}^2$. The distance travelled by truck before coming to rest is

A ball is thrown vertically upwards with an initial velcoity $u$ reaches maximum height in 5 s . The ratio of distance travelled by the ball in the 2nd and 7th second is (assume, $g=10 \mathrm{~m} / \mathrm{s}^2$ )

A ball is thrown straight upward from ground with a speed of $20 \mathrm{~m} / \mathrm{s}$. The ball was caught on its way down at a point 5 m above the ground. The time taken by the ball during entire trip is (assume, $g=10 \mathrm{~m} / \mathrm{s}^2$ )

A motor-bike starts from rest, attains a velocity of 10 $\mathrm{m} / \mathrm{s}$ with an acceleration of $0.5 \mathrm{~m} / \mathrm{s}^2$, travels 10 km with this uniform velocity and then comes to halt with a uniform deceleration of $0.2 \mathrm{~m} / \mathrm{s}^2$. The total time of travel is

A particle is moving along the $Y$-axis. The position of the particle from the origin as a function of time $(t)$ is given as $y(t)=10 t e^{-2 t}$. How far is the particle from the origin when it stops momentarily? ( $y$ is given in units of metre and $t$ is in units of second)

A car travelling at $15 \mathrm{~m} / \mathrm{s}$ overtake another car travelling at $10 \mathrm{~m} / \mathrm{s}$. Assuming, each car is 4 m long. What is the time taken during the overtake?