Motion in a Straight LineNEET Physics · Class 11 · NCERT Chapter 2

A particle starts from rest with a constant acceleration of $2{\text{m/s}}^2$. Its velocity after $5\text{s}$ is:

A car starting from rest accelerates uniformly at $4{\text{m/s}}^2$. The distance it covers in the $5$th second of its motion is:

A body moves in a straight line with constant acceleration. It covers $20\text{m}$ in the $4$th second and $30\text{m}$ in the $9$th second. The acceleration is:

A particle moves in a straight line with initial velocity $10\text{m/s}$ and constant acceleration $2{\text{m/s}}^2$. The distance covered in $4\text{s}$ is:

A particle moving with velocity $u$ on a straight line has uniform retardation $a$ until it stops. The total distance covered is:

A body moving with velocity $u$ doubles its speed after travelling a distance $s$ under uniform acceleration $a$. Then $a$ equals:

A stone is dropped from a height of $80\text{m}$. The time it takes to reach the ground (take $g=10{\text{m/s}}^2$) is:

A ball is thrown vertically upward with a velocity of $20\text{m/s}$. The maximum height reached (take $g=10{\text{m/s}}^2$) is:

A body is thrown vertically upward with velocity $u$. The total time it spends in the air before returning to the launch point is:

A stone is dropped from a height of $45\text{m}$ from a cliff. Take $g=10{\text{m/s}}^2$. Its velocity just before hitting the ground is:

A ball is dropped from the top of a tower. In the last second of its fall, it covers a distance of $25\text{m}$. The height of the tower (take $g=10{\text{m/s}}^2$) is:

Two stones are dropped from a cliff one after the other with a time gap of $1\text{s}$. After how many seconds (from the dropping of the first stone) is the gap between them $25\text{m}$? Take $g=10{\text{m/s}}^2$.

A car covers the first half of a journey at $60\text{km/h}$ and the second half at $40\text{km/h}$. Its average speed for the whole journey is:

A particle covers the first one-third of a distance at $30\text{km/h}$, the second one-third at $40\text{km/h}$ and the last one-third at $60\text{km/h}$. The average speed is:

A man walks $3\text{km}$ east in $1\text{h}$ and then $4\text{km}$ west in $1\text{h}$. His average speed and average velocity are:

The position of a particle is given by $x(t)=3t^2+2t+5$ (in metres, $t$ in seconds). Its instantaneous velocity at $t=2\text{s}$ is:

The displacement of a particle in time $t$ is given by $x=4t^2-2t+1$ (SI units). Its initial velocity is:

The position of a particle moving in a straight line is given by $x(t)=t^3-6t^2+9t+2$ (SI units). Its acceleration at $t=2\text{s}$ is:

A car accelerates uniformly from $20\text{m/s}$ to $40\text{m/s}$ in $5\text{s}$. Its acceleration is:

A car moving at $108\text{km/h}$ comes to rest in $5\text{s}$ after the brakes are applied. The retardation is:

In a velocity-time graph, the area enclosed between the graph and the time axis represents:

On a position-time graph, the slope at any instant represents:

The velocity-time graph of a particle is a straight line with positive slope passing through the origin. The motion is:

A particle moves along the x-axis with velocity-time graph as a triangle of base $10\text{s}$ and height $20\text{m/s}$ (above the time axis). The displacement covered is:

A position-time graph for a particle is a parabola passing through the origin opening upward. The motion is:

Two cars A and B move on a straight road in the same direction with speeds $60$ and $40\text{km/h}$ respectively. The relative velocity of A with respect to B is:

Two trains, $100\text{m}$ and $200\text{m}$ long, move in opposite directions at $20\text{m/s}$ and $30\text{m/s}$ respectively on parallel tracks. The time to cross each other is:

A boy on a moving bus drops a ball from his hand. To an observer on the ground, the ball appears to:

A man walks $4\text{m}$ north and then $3\text{m}$ east. His displacement and distance are respectively:

A person completes one round of a circular track of radius $R$. His displacement is:

A particle moves along the x-axis. Its position at $t=0$ is $x=+5\text{m}$ and at $t=4\text{s}$ is $x=-3\text{m}$. Its displacement during this interval is:

A car moving at $20\text{m/s}$ stops in $40\text{m}$ on a wet road. On the same road, the same car moving at $30\text{m/s}$ would stop in: