Motion in a Plane PYQs

Two cities X and Y are connected by a regular bus service with a bus leaving in either direction every T min. A girl is driving scooty with a speed of 60 km/h in the direction X to Y notices that a bus goes past her every 30 minutes in the direction of her motion, and every 10 minutes in the opposite direction. Choose the correct option for the period T of service and the speed (assumed constant) of the buses.

A particle moving with uniform speed in a circular path maintains:

A bullet is fired from a gun at the speed of 280 m s⁻¹ in the direction 30° above the horizontal. The maximum height attained by the bullet is : (g = 9.8 m s⁻², sin 30° = 0.5)

A ball is projected with a velocity 10 ms⁻¹, at an angle of 60° with the vertical direction. Its speed at the highest point of its trajectory will be:

A car starts from rest and accelerates at 5 m/s². At t = 4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t = 6 s? (Take g = 10 m/s²)

A particle moving in a circle of radius $R$ with a uniform speed takes a time $T$ to complete one revolution. If this particle were projected with the same speed at an angle $\theta$ to the horizontal, the maximum height attained by it equals $4R$. The angle of projection, $\theta$, is given by:

A particle starting from rest, moves in a circle of radius r. It attains a velocity of V₀ m/s in the nᵗʰ round. Its angular acceleration will be:

Two bullets are fired horizontally and simultaneously towards each other from rooftops of two buildings $100\ \text{m}$ apart and of same height $200\ \text{m}$, with the same velocity of $25\ \text{m/s}$ $(g = 10\ \text{m/s}^2)$. When and where will the two bullets collide?

The $x$ and $y$ coordinates of the particle at any time are $x = 5t - 2t^2$ and $y = 10t$ respectively, where $x$ and $y$ are in meters and $t$ in seconds. The acceleration of the particle at $t = 2\ \text{s}$ is

Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time $t_1$. On another day, if she remains stationary on the moving escalator, then the escalator takes her up in time $t_2$. The time taken by her to walk up on the moving escalator will be:

In the given figure, $a = 15\ \text{m/s}^2$ represents the total acceleration of a particle moving in the clockwise direction in a circle of radius $R = 2.5\ \text{m}$ at a given instant of time. The speed of the particle is:

If vectors $$\vec{A}=\cos\omega t\,\hat{i}+\sin\omega t\,\hat{j}$$ $$\vec{B}=\cos\left(\frac{\omega t}{2}\right)\hat{i}+\sin\left(\frac{\omega t}{2}\right)\hat{j}$$ are functions of time, then the value of $t$ at which they are orthogonal to each other is:

A projectile is fired from the surface of the earth with a velocity of $5\,\text{m s}^{-1}$ and angle $\theta$ with the horizontal. Another projectile fired from another planet with a velocity of $3\,\text{m s}^{-1}$ at the same angle follows a trajectory which is identical to the trajectory of the projectile fired from earth. The value of the acceleration due to gravity on the planet is (in $\text{m s}^{-2}$) given $g = 9.8\,\text{m s}^{-2}$.

A particle is moving such that its position coordinates $(x, y)$ are $(2\,\text{m}, 3\,\text{m})$ at time $t = 0$, $(6\,\text{m}, 7\,\text{m})$ at time $t = 2\,\text{s}$ and $(13\,\text{m}, 14\,\text{m})$ at time $t = 5\,\text{s}$. Average velocity vector $\left(\vec{v}_{\text{av}}\right)$ from $t = 0$ to $t = 5\,\text{s}$ is:

The velocity of a projectile at the initial point $A$ is $(2\hat{i} + 3\hat{j})\ \mathrm{m/s}$. Its velocity (in $\mathrm{m/s}$) at point $B$ is

A projectile is fired at an angle of 45° with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is:

The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is

A particle moves in the $xy$-plane according to the rule $$x = a\sin\omega t$$ $$y = a\cos\omega t$$ The particle follows

A bus is moving with a speed of $10\,\mathrm{m\,s^{-1}}$ on a straight road. A scooterist wishes to overtake the bus in $100\,\mathrm{s}$. If the bus is at a distance of $1\,\mathrm{km}$ from the scooterist, with what speed should the scooterist chase the bus?

A particle of mass m is projected with velocity v making an angle of 45° with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be

An airplane is moving horizontally with speed 100 m/s at a height of 2000 m from the ground. A small object is detached from it and strikes the ground. Calculate the angle from the vertical with which it strikes the ground.

For angles of projection of a projectile at angles (45° − θ) and (45° + θ), the horizontal ranges described by the projectile are in the ratio of:

The vectors $\vec{A}$ and $\vec{B}$ are such that a: The angle between the two vectors is:

If a vector $$2\hat{i}+3\hat{j}+8\hat{k}$$ is perpendicular to the vector $$4\hat{j}-4\hat{i}+\alpha\hat{k}$$ then the value of $\alpha$ is:

Two boys are standing at the ends A and B of a ground, where $AB = a$. The boy at B starts running in a direction perpendicular to $AB$ with velocity $v_1$. The boy at A starts running simultaneously with velocity $v$ and catches the other boy in a time $t$. Then $t$ is:

If the angle between the vectors $\vec{A}$ and $\vec{B}$ is $\theta$, the value of the product $(\vec{B} \times \vec{A}) \cdot \vec{A}$ is equal to: