iCON Education - Motion in a Straight Line AP-EAPCET Test

A ball projected vertically upwards with a velocity ' $V$ ' passes through a point $P$ in its upward journey in a time of ' $x$ ' seconds. From there, the time in which the ball again passes through the same point $P$ is

$2\left(\frac{v}{g}-x\right)$

Two smooth inclined planes $A$ and $B$ each of height 20 m have angles of inclination $30^{\circ}$ and $60^{\circ}$ respectively. If $t_1$ and $t_2$ are respectively the times taken by two blocks to reach the bottom of the planes $A$ and $B$ from the top, then $t_1-t_2=$ (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

$\frac{\sqrt{3}-1}{\sqrt{3}} \mathrm{~s}$

$3(\sqrt{3}-1) \mathrm{s}$

$4\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right) \mathrm{s}$

$(3 \sqrt{3}-2) \mathrm{s}$

If the displacement ( $s$ in metre) of a moving particle in terms of time $(t$ in second $) s=t^3-6 t^2+18 t+9$, then the minimum velocity attained by the particle is

$29 \mathrm{~ms}^{-1}$

$5 \mathrm{~ms}^{-1}$

$6 \mathrm{~ms}^{-1}$

$12 \mathrm{~ms}^{-1}$

The displacement $(x)$ and time $(t)$ graph of a particle moving along a straight line is shown in the figure. The average velocity of the particle in the time of 10 s is

AP EAPCET 2025 - 26th May Evening Shift Physics - Motion in a Straight Line Question 2 English

$2 \mathrm{~ms}^{-1}$

$4 \mathrm{~ms}^{-1}$

$6 \mathrm{~ms}^{-1}$

$8 \mathrm{~ms}^{-1}$

A body starts from rest with uniform acceleration and its velocity at a time of ' $n$ ' seconds is ' $v$ '. The total displacement of the body in the $n$th and $(n-1)$ th seconds of its motion is

$\frac{2 v(n+1)}{n}$

$\frac{2 v(n-1)}{n}$

If a stone thrown vertically upwards from a bridge with an initial velocity of $5 \mathrm{~ms}^{-1}$, strikes the water below the bridge in a time of 3 s , then the height of the bridge above the water surface is (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

A particle moving along a straight line covers the first half of the distance with a speed of $3 \mathrm{~ms}^{-1}$, the other half of the distance is covered in two equal time intervals with speeds of $4.5 \mathrm{~ms}^{-1}$ and $7.5 \mathrm{~ms}^{-1}$ respectively, then the average speed of particle during the motion is

$4.0 \mathrm{~ms}^{-1}$

$5.0 \mathrm{~ms}^{-1}$

$5.5 \mathrm{~ms}^{-1}$

$4.8 \mathrm{~ms}^{-1}$

If the distance travelled by a freely falling body in the last but one second of its motion is 5 m , then the last second is

(Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

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

If a car travels $40 \%$ of the total distance with a speed $v_1$ and the remaining distance with a speed $v_2$, then average speed of the car is

$\frac{1}{2} \sqrt{v_1 v_2}$

$\frac{2 v_1 v_2}{v_1+v_2}$

$\frac{5 v_1 v_2}{3 v_1+2 v_2}$

A diving board is at at height of $h$ from the water surface. A swimmer standing on this board thrown a stone vertically upward with a velocity $16 \mathrm{~ms}^{-1}$. It reaches the water surface in a time of 5 s . In the next 0.2 s the diver can hear the sound from water surface. The speed of sound is (acceleration due to gravity $g=10 \mathrm{~ms}^{-2}$ )

$450 \mathrm{~ms}^{-1}$

$225 \mathrm{~ms}^{-1}$

$200 \mathrm{~ms}^{-1}$

$275 \mathrm{~ms}^{-1}$

A person walks up a stalled escalator in 90 s. When standing on the same moving escalator, he reached in 60s. The time it would take him to walk up the moving escalator will be

A particle starts from rest and moves in a straight line. It travels a distance $2 L$ with uniform acceleration and then moves with a constant velocity a further distance of $L$. Finally, it comes to rest after moving a distance of $3 L$ under uniform retardation. Then, the ratio of average speed to the maximum speed $\left(\frac{v}{v_m}\right)$ of the particle is

The velocity of a particle is given by the equation $v(x)=3 x^2-4 x$, where $x$ is the distance covered by the particle. The expression for its acceleration is

$6\left(3 x^2-4 x\right)$

$\left(3 x^2-4 x\right)(6 x-4)$

The acceleration of a particle which moves along the positive $X$-axis varies with its position as shown in the figure. If the velocity of the particle is $0.8 \mathrm{~ms}^{-1}$ at $x=0$ , then its velocity at $x=1.4 \mathrm{~m}$ is $\left(\right.$ in $\left.\mathrm{ms}^{-1}\right)$

AP EAPCET 2024 - 21th May Evening Shift Physics - Motion in a Straight Line Question 14 English

An object projected upwards from the foot of a tower. The object crosses the top of the tower twice with an interval of 8 s and the object reaches foot after 16 s . The height of the tower is $\left(g=10 \mathrm{~ms}^{-2}\right)$

The relation between time $t$ and displacement $x$ is $t=\alpha x^2+\beta x$, where $\alpha$ and $\beta$ are constants. If $v$ is the velocity, the retardation is

$2 \alpha v^3 \beta^2$

$2 \alpha \beta v^3$

A body starting from rest moving with an acceleration of $\frac{5}{4} \mathrm{~ms}^{-2}$. The distance travelled by the body in the third second is

A car travelling at 80 kmph can be stopped at a distance of 60 m by applying brakes. If the same car travels at 160 kmph and the same braking force is applied, the stopping distance is

A student is at a distance 16 m from a bus when the bus begins to move with a constant acceleration of $9 \mathrm{~m} \mathrm{~s}^{-2}$. The minimum velocity with which the student should run. towards the bus so as the catch it is $\alpha \sqrt{2} \mathrm{~ms}^{-1}$. The value of $\alpha$ is

An object moving along $X$-axis with a uniform acceleration has velocity $\mathbf{v}=\left(12 \mathrm{cms}^{-1}\right) \hat{\mathbf{i}}$ at $x=3 \mathrm{~cm}$. After 2 s if it is at $x=-5 \mathrm{~cm}$, then its acceleration is

$\mathbf{a}=\left(-16 \mathrm{~cms}^{-2}\right) \hat{i}$

$\mathbf{a}=\left(11 \mathrm{~cms}^{-2}\right) \hat{i}$

$\mathbf{a}=\left(-11 \mathrm{~cms}^{-2}\right) \hat{i}$

$\mathbf{a}=\left(8 \mathrm{~cms}^{-2}\right) \hat{i}$

$y=\left(P t^2-Q t^3\right) \mathrm{~m}$ is the vertical displacement of a ball which is moving in vertical plane. Then the maximum height that the ball can reach is

$\frac{27 P^3}{4 Q^2}$

$\frac{4 Q^2}{27 P^3}$

$\frac{4 P^3}{27 Q^2}$

$\frac{27 Q^2}{4 P^3}$

A car covers a distance at speed of $60 \mathrm{~km} \mathrm{~h}^{-1}$. It returns and comes back to the original point moving at a speed of $v$. If the average speed for the round trip is $48 \mathrm{~kmh}^{-1}$, then the magnitude of $v$ is

$40 \mathrm{~km} \mathrm{~h}^{-1}$

$36 \mathrm{~km} \mathrm{~h}^{-1}$

$44 \mathrm{~km} \mathrm{~h}^{-1}$

$32 \mathrm{~km~h}^{-1}$

An object is moving with a uniform acceleration which is parallel to its instantaneous direction of motion. The displacement-velocity graph of this object is

The displacement of a particle starting from rest at $t=0$ is given by $s=9 t^2-2 t^3$. The time in seconds at which the particle will attain zero velocity is

Two cars A and B are moving with a velocity of 30 km/h in the same direction. They are separated by 10 km. The speed of another car C moving in the opposite direction, if it meets these two cars at an interval of eight minutes is

An object travelling at a speed of 36 km/h comes to rest in a distance of 200 m after the brakes were applied. The retardation produced by the brakes is

A ball is projected upwards. Its acceleration at the highest point is

such as cannot be predicted

Which of the following decreases, in motion on a straight line, with constant retardation?