Motion in a Straight Line

If the instantaneous velocity of a particle is constant, its path must be:

The instantaneous velocity of an object moving in a straight line is zero when:

Instantaneous velocity is also known as:

If the slope of the tangent to an $x-t$ graph is positive, the instantaneous velocity is:

If the position of an object is $x = 4 \cos(2t)$, what is the instantaneous velocity at $t = \frac{\pi}{4}\ s$?

A particle moves such that $x = t^2 + 2t$. The average velocity between $t = 0$ and $t = 2\ s$ is equal to the instantaneous velocity at $t = T$. Find $T$.

The displacement of an object is $x = (2t - 1)^3$. What is its instantaneous velocity at $t = 1\ s$?

A particle moves along a straight line such that $x = t^3/3 - 2t^2 + 3t + 4$. The velocity of the particle is zero at $t$ equal to:

A body moves in a straight line with $v = 4 + 2t$. The distance covered by the body in the first $2$ seconds is:

A particle moves such that its velocity $v = k x^2$. The acceleration of the particle when $x = 2$ is:

A person $P$ is moving at $5\text{ km/h}$ and $Q$ is moving at $5\text{ km/h}$ in opposite directions. The relative velocity of $P$ with respect to $Q$ is:

If $v_A = 20\text{ m/s}$ and $v_B = -10\text{ m/s}$, find $v_{AB}$.

Relative velocity of ground with respect to a person running at $5\text{ m/s}$ East is:

If two bodies move in the same direction with $v_1$ and $v_2$, the distance between them changes at the rate of:

Two bodies are moving in the same direction with velocities $v_1$ and $v_2$. The distance between them is increasing at the rate of $4 \text{ m/s}$. If they move in opposite directions, the distance between them increases at $10 \text{ m/s}$. The values of $v_1$ and $v_2$ are:

A projectile is fired with a velocity $v$ at an angle $\theta$ to the horizontal. The relative velocity of the projectile with respect to another projectile fired with the same velocity and angle from the same point is:

A man is running on a flat car with a velocity of $5 \text{ m/s}$ relative to the car. The car is moving with a velocity of $15 \text{ m/s}$ relative to the ground in the same direction. What is the velocity of the man relative to the ground?

Two cars are moving on a straight road. Car $A$ is at $x=0$, $t=0$ moving at $10\text{ m/s}$ with acceleration $2\text{ m/s}^2$. Car $B$ is at $x=100\text{ m}$, $t=0$ moving at $20\text{ m/s}$ with acceleration $1\text{ m/s}^2$. They are moving in the same direction. At what time will they meet?

Two objects $A$ and $B$ move with velocities $10\text{ m/s}$ and $20\text{ m/s}$ in opposite directions. The relative velocity of $A$ with respect to $B$ is:

A particle $A$ is moving with velocity $v_1$ and another particle $B$ with velocity $v_2$. If the angle between them is $\theta$, the magnitude of relative velocity of $A$ with respect to $B$ is:

For an object starting from rest, the displacement in $3$ seconds is how many times the displacement in $1$ second?

A ball is dropped from a height of $5\ m$. What is its velocity just before hitting the ground? ($g = 10\ m/s^2$)

In a position-time ($x-t$) graph, a parabola opening upwards represents:

A scooter moving at $10\ m/s$ is stopped by brakes in $5\ m$. What is the acceleration?

A driver of a car traveling at $15\ m/s$ applies the brakes when he sees a red light. If his reaction time is $0.2\ s$ and the brakes cause a constant retardation of $6\ m/s^2$, what is the total distance traveled before the car stops?

A particle starts from rest with constant acceleration. The ratio of the average velocity during the first $n$ seconds to the final velocity at the end of $n$ seconds is:

A body is moving with uniform acceleration and describes $40\ m$ in the $3^{rd}$ second and $120\ m$ in the $5^{th}$ second. What is the distance described in the $10^{th}$ second?

A stone is dropped from a height $h$. In the last second of its motion, it covers $0.45h$. Find the value of $h$. (Take $g = 10\, m/s^2$)

A point particle starting from rest has a velocity that increases linearly with time such that $v = kt$ where $k = 2\, m/s^2$. The distance covered in the interval $t = 2\, s$ to $t = 4\, s$ is:

A body moves from rest with uniform acceleration and travels $S\, m$ in $n$ seconds. The distance traveled in the last $2$ seconds is:

If an object is moving with constant negative velocity, its position-time graph will be:

The slope of a velocity-time graph gives:

In a velocity-time graph, a horizontal line at $v = 0$ means the object is:

If the position-time graph is a parabola opening upwards, the motion is:

A particle starts from rest and moves with constant acceleration. Which graph correctly represents the relationship between its displacement ($s$) and time ($t$)?

Study the velocity-time graph. What is the total distance traveled by the object in 5 seconds?

For the given position-time graph, what is the instantaneous velocity at $t = 2$ s?

A particle's motion is described by the velocity-time graph below. The area under the graph from $t=0$ to $t=2\text{ s}$ is $A_1$ and from $t=2$ to $t=5\text{ s}$ is $A_2$. If $A_1 = 10\text{ m}$ (positive) and $A_2 = -15\text{ m}$ (negative), find the total distance traveled and the final displacement at $t=5\text{ s}$ from the start.

The displacement-time graph of a ball bouncing elastically on a hard floor is shown. If the peak of each bounce is represented by the vertices of the parabolas, and the floor is at $x=0$, which of the following represents the velocity-time graph for two consecutive bounces?

A particle moves in a straight line such that its velocity $v$ changes with displacement $s$ as shown in the $v-s$ graph (a straight line from $(0, 10)$ to $(20, 0)$). What is the magnitude of the acceleration of the particle when its displacement is $10\text{ m}$?

A ball is thrown upwards. While it is rising, the acceleration due to gravity is acting:

The direction of the acceleration vector is determined by the direction of:

What is the acceleration of a vehicle that maintains a constant speed of $60\ km/h$ on a straight road?

Acceleration is a rate of change of a rate of change of:

A body is thrown vertically upwards. If it travels a distance $h$ in the $n^{th}$ second, then the distance travelled in the $(n+1)^{th}$ second is:

If a body travels half of its total path in the last second of its fall from rest, the time of fall is:

The velocity of a particle is $v = v_0 + gt + ft^2$. If its position is $x=0$ at $t=0$, then its displacement after time $t = 1\, s$ is:

The displacement of a particle is given by $y = a + bt + ct^2 - dt^4$. The initial velocity and initial acceleration are respectively:

If a body travels half of its total path in the last second of its fall from rest, find the total height of its fall ($g = 9.8\text{ m/s}^2$):

Two bodies $A$ and $B$ are thrown simultaneously, $A$ vertically upwards with $20\text{ m/s}$ from the ground and $B$ vertically downwards from a height of $40\text{ m}$ with the same speed. Where will they meet?