Motion In A Straight Line - Mechanics - Physics Previous Year Questions

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 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

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

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

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 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 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, $g=10 \mathrm{~m} / \mathrm{s}^2$ )