Q.A particle starts from the origin at $t = 0\ \text{s}$ with a velocity of $10.0\,\hat{j}\ \text{m/s}$ and moves in the $x$-$y$ plane with a constant acceleration of $(8.0\,\hat{i} + 2.0\,\hat{j})\ \text{m s}^{-2}$.

Showing the 12 most recent of 25 on this concept.

1. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.A particle has initial velocity $2\hat{i}+3\hat{j}\ ms^{-1}$ and acceleration $0.8\hat{i}+0.6\hat{j}\ ms^{-2}$. Its velocity after 5 sec is (in $ms^{-1}$) (A) $6\sqrt{2}$ (B) 6 (C) $7\sqrt{2}$ (D) 7

   ›Reveal solutionSolution

   Straightforward kinematics with vectors: add the acceleration's contribution to each component of velocity separately, then find the magnitude.

   Concept and Intuition

   Since acceleration is constant here, each Cartesian component of velocity evolves independently and linearly in time: $v_x(t) = v_{0x}+a_x t$ and $v_y(t) = v_{0y}+a_y t$. There's no need to find a resultant acceleration vector's magnitude/direction first — just apply the equation component-by-component and combine at the end.

   Step-by-Step Solution

   1. Given $\vec v_0 = 2\hat i + 3\hat j\ ms^{-1}$ and $\vec a = 0.8\hat i + 0.6\hat j\ ms^{-2}$, and $t=5\,s$.
   2. $x$-component: $v_x = 2 + 0.8(5) = 2+4 = 6\ ms^{-1}$.
   3. $y$-component: $v_y = 3 + 0.6(5) = 3+3 = 6\ ms^{-1}$. …

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2. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.A particle initially at rest starts moving from origin along X-axis with velocity 'V' that varies as, $V = 2\sqrt{x}\ \text{ms}^{-1}$. The acceleration of the particle in $\text{ms}^{-2}$ is (A) $2$ (B) $2\sqrt{2}$ (C) $\dfrac{2}{\sqrt{2}}$ (D) Zero

   ›Reveal solutionSolution

   Using $a = v\,dv/dx$ on $v = 2\sqrt{x}$ gives a constant acceleration of

   $2\ \text{ms}^{-2}$. Answer: (A).

   Concept and Intuition

   When velocity is given as a function of position (not time), the cleanest way

   to get acceleration is the identity $a = \dfrac{dv}{dt} = \dfrac{dv}{dx} \cdot \dfrac{dx}{dt} = v\dfrac{dv}{dx}$. This avoids needing to first invert

   $v(x)$ into $x(t)$.

   Step-by-Step Solution

   1. Given $v = 2\sqrt{x} = 2x^{1/2}$.
   2. Differentiate w.r.t. $x$: $\dfrac{dv}{dx} = 2\cdot\dfrac{1}{2}x^{-1/2} = \dfrac{1}{\sqrt{x}}$.
   3. Apply $a = v\dfrac{dv}{dx} = \left(2\sqrt{x}\right)\cdot \left(\dfrac{1}{\sqrt{x}}\right) = 2$.
   4. The $\sqrt{x}$ terms cancel exactly, so acceleration is the constant …

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3. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.A velocity-time graph is drawn for two different objects. They make $30°$ and $45°$ with the time axis. Then the ratio of their accelerations, $a_1 : a_2$ is (A) $1 : 2$ (B) $2 : 3$ (C) $\sqrt{3} : 1$ (D) $1 : \sqrt{3}$

   ›Reveal solutionSolution

   Acceleration equals the slope of the v–t graph, so $a_1:a_2=\tan30°:\tan45°=1:\sqrt3$.

   Concept and Intuition

   On a velocity-time graph, the instantaneous acceleration is $\dfrac{dv}{dt}$, which is exactly the slope of the graph — and slope equals the tangent of the angle the line makes with the time axis.

   Step-by-Step Solution

   1. $a_1=\tan30°=\dfrac{1}{\sqrt3}$
   2. $a_2=\tan45°=1$
   3. Ratio:

   $\frac{a_1}{a_2}=\frac{1/\sqrt3}{1}=\frac{1}{\sqrt3}$

   4. So $a_1:a_2=1:\sqrt3$. …

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4. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.A particle of charge 'q' and mass 'm' starts moving from the origin under the action of electric field, $\vec{E} = E_0\hat{i}$ with a velocity $v = v_0\hat{j}$. The time taken to increase its velocity to $\dfrac{\sqrt{5}}{2}v_0$ is (A) $\dfrac{mv_0}{qE_0}$ (B) $\dfrac{mv_0}{2qE_0}$ (C) $\dfrac{\sqrt{3}\,mv_0}{2qE_0}$ (D) $\dfrac{\sqrt{5}\,mv_0}{2qE_0}$

   ›Reveal solutionSolution

   The field only accelerates the particle along $x$; the $y$-velocity stays fixed at $v_0$, so the target speed condition fixes $v_x=v_0/2$, giving $t=\dfrac{mv_0}{2qE_0}$.

   Concept and Intuition

   The electric field acts only along $\hat i$, so it produces a constant acceleration $a=qE_0/m$ purely in the $x$-direction — exactly like projectile motion, where the $y$-component of velocity is completely unaffected by a force acting only along $x$. The particle's speed at any time is found from both components combined via Pythagoras.

   Step-by-Step Solution

   1. Acceleration: $a=\dfrac{qE_0}{m}$, directed along $x$.
   2. Velocity components at time $t$: $v_x(t)=at=\dfrac{qE_0}{m}t$, and $v_y(t)=v_0$ (unchanged, no force along $y$).
   3. Required speed: $v=\dfrac{\sqrt5}{2}v_0 \Rightarrow v^2=v_x^2+v_y^2=\dfrac54v_0^2$. …

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5. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.A particle moves with velocity $v = at - bt^2$ where $a$ and $b$ are constants. The acceleration becomes zero at (A) $t = a/b$ (B) $t = a/2b$ (C) $t = 2a/b$ (D) $t = b/a$

   ›Reveal solutionSolution

   Differentiate the given velocity to get acceleration, then find when it vanishes; $t=a/2b$.

   Concept and Intuition

   Acceleration is the instantaneous rate of change of velocity. Given $v(t)$ as a function of time, $a(t)=dv/dt$. Setting $a(t)=0$ finds the instant the velocity is momentarily stationary (an extremum of $v$).

   Step-by-Step Solution

   1. $v = at - bt^2$.
   2. $a_{acc} = \dfrac{dv}{dt} = a - 2bt$.
   3. Set $a_{acc}=0$: $a - 2bt = 0 \Rightarrow t = \dfrac{a}{2b}$.

   Common Mistakes …

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6. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.A particle is moving along a straight line such that its velocity is increasing at 5 ms$^{-1}$ per meter. When its velocity becomes 20 ms$^{-1}$ its acceleration is (A) 50 ms$^{-2}$ (B) 75 ms$^{-2}$ (C) 100 ms$^{-2}$ (D) 10 ms$^{-2}$

   ›Reveal solutionSolution

   A velocity-vs-position relation is turned into acceleration using $a = v\,dv/dx$. Answer: (C) 100 ms$^{-2}$.

   Concept and Intuition

   When velocity is given as a function of position rather than time, the direct route to acceleration is the chain rule: $a = \dfrac{dv}{dt} = \dfrac{dv}{dx}\cdot\dfrac{dx}{dt} = v\dfrac{dv}{dx}$. This is the standard trick for "velocity increases so much per metre" problems — you never need to know how it depends on time at all.

   Step-by-Step Solution

   1. The statement "velocity is increasing at 5 ms$^{-1}$ per metre" means $\dfrac{dv}{dx} = 5\ \text{s}^{-1}$ (constant).
   2. Use $a = v\dfrac{dv}{dx}$.
   3. At the instant $v = 20\ \text{ms}^{-1}$: $a = 20 \times 5 = 100\ \text{ms}^{-2}$. …

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7. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.A particle crossing the origin at time $t=0$, moves in the xy-plane with a constant acceleration 'a' in y-direction. If the equation of motion of the particle is $y = bx^2$ (where b is a constant), then its velocity component in the x-direction is (A) $\sqrt{\dfrac{2b}{a}}$ (B) $\sqrt{\dfrac{a}{2b}}$ (C) $\sqrt{\dfrac{a}{b}}$ (D) $\sqrt{\dfrac{b}{a}}$

   ›Reveal solutionSolution

   Matching the parabolic path $y=bx^2$ to the kinematic equations of constant-velocity $x$-motion and constant-acceleration $y$-motion gives $v_x=\sqrt{a/2b}$.

   Concept and Intuition

   This is projectile-like motion: since acceleration acts only along $y$, the $x$-motion is uniform (constant velocity), while $y$-motion starts from rest and accelerates uniformly. The given trajectory equation $y=bx^2$ must be consistent with both these kinematic equations at every instant $t$, which pins down $v_x$ in terms of $a$ and $b$.

   Step-by-Step Solution

   1. Since there's no acceleration along $x$: $x=v_x t$ (constant velocity $v_x$).
   2. Since acceleration $a$ acts along $y$ starting from rest at the origin: $y=\dfrac12 a t^2$.
   3. The trajectory is given as $y=bx^2$. Substitute $x=v_xt$: $y=b(v_xt)^2=bv_x^2t^2.$ …

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8. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.For a particle moving in x-y plane, if at any instant of time 't', (in second) its displacements (in metre) are $x = 2t^2 - t$ and $y = 4t^2 - 4t$, then the velocity of the particle at a time $t = 1\ \mathrm{s}$ is (A) $3\ \mathrm{m\,s^{-1}}$ (B) $7\ \mathrm{m\,s^{-1}}$ (C) $1\ \mathrm{m\,s^{-1}}$ (D) $5\ \mathrm{m\,s^{-1}}$

   ›Reveal solutionSolution

   Differentiate $x(t)$ and $y(t)$ to get the velocity components, then combine them via the Pythagorean theorem to get the speed. Answer: (D) 5 m/s.

   Concept and Intuition

   For 2D motion described by position components $x(t)$ and $y(t)$, the instantaneous velocity components are obtained by differentiating each with respect to time: $v_x = dx/dt$, $v_y = dy/dt$. The magnitude (speed) of the velocity vector is then found via $|\vec v| = \sqrt{v_x^2+v_y^2}$, exactly as with any 2D vector.

   Step-by-Step Solution

   1. Given: $x = 2t^2 - t$, so $v_x = \dfrac{dx}{dt} = 4t - 1$.
   2. Given: $y = 4t^2 - 4t$, so $v_y = \dfrac{dy}{dt} = 8t - 4$.
   3. Evaluate at $t = 1\ \mathrm{s}$: $v_x = 4(1)-1 = 3\ \mathrm{m/s}$; $v_y = 8(1)-4 = 4\ \mathrm{m/s}$.
   4. Compute the speed: $|\vec v| = \sqrt{v_x^2 + v_y^2} = \sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5\ \mathrm{m/s}$. …

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9. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.The displacement (s) - time (t) graphs of two bodies A and B are shown in the figure. The ratio of the velocities of the two bodies A and B is [FIGURE] (a graph with t on the vertical axis and s on the horizontal axis; two straight lines drawn from the origin — line B making a 30° angle with the vertical t-axis, and line A making a 45° angle with the horizontal s-axis) (A) $1:\sqrt{3}$ (B) $2:\sqrt{3}$ (C) $\sqrt{3}:1$ (D) $\sqrt{3}:2$

   ›Reveal solutionSolution

   On this swapped-axis plot ($t$ vertical, $s$ horizontal), $v=\frac{\Delta s}{\Delta t}$: for A ($45^\circ$ to the $s$-axis) $v_A=1$; for B ($30^\circ$ to the $t$-axis) $v_B=\tan30^\circ=\frac{1}{\sqrt3}$; so $v_A:v_B=\sqrt3:1$ -> option (C).

   The axes are unusual: time $t$ is on the vertical axis and displacement $s$ on the horizontal axis. Velocity is still $v=\dfrac{\Delta s}{\Delta t}$ = (horizontal change) $/$ (vertical change).

   Body A - line at $45^\circ$ to the horizontal $s$-axis. Measuring from the horizontal, $\tan45^\circ=\dfrac{\Delta t}{\Delta s}=1$, so

   $v_A=\frac{\Delta s}{\Delta t}=1.$ …

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10. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
    Q.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 (A) $2\alpha v^3 \beta^2$ (B) $2\alpha\beta v^3$ (C) $-2\beta v^3$ (D) $2\alpha v^3$

    ›Reveal solutionSolution

    Differentiating $t=\alpha x^2+\beta x$ twice (via $v=dx/dt$ and $a=v\,dv/dx$) gives the deceleration as $2\alpha v^3$ — (D).

    Concept and Intuition

    When $t$ is given as a function of $x$ (rather than the usual $x(t)$), it's often easiest to first get $v=dx/dt$ as a function of $x$ by inverting $dt/dx$, and then use the identity $a=v\dfrac{dv}{dx}$ to find acceleration without ever needing an explicit $x(t)$.

    Step-by-Step Solution

    1. $t=\alpha x^2+\beta x\Rightarrow\dfrac{dt}{dx}=2\alpha x+\beta$.
    2. Since $v=\dfrac{dx}{dt}$, we have $v=\dfrac{1}{dt/dx}=\dfrac1{2\alpha x+\beta}$.
    3. Differentiate $v$ with respect to $x$: $\dfrac{dv}{dx}=-\dfrac{2\alpha}{(2\alpha x+\beta)^2}=-2\alpha v^2$ (recognizing $(2\alpha x+\beta)^{-2}=v^2$).
    4. Acceleration: $a=v\dfrac{dv}{dx}=v\cdot(-2\alpha v^2)=-2\alpha v^3$. …

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11. AP EAPCET 2023Set ap-2023-05-22-FN1 markMCQ
    Q.The distance s (in m) travelled by a particle is given by the expression $4s = 5t^{2} + 8t$, here t is time in seconds. The acceleration of the particle is (A) $5\ \mathrm{ms^{-2}}$ (B) $2.5\ \mathrm{ms^{-2}}$ (C) $1.25\ \mathrm{ms^{-2}}$ (D) $2\ \mathrm{ms^{-2}}$

    ›Reveal solutionSolution

    A kinematics differentiation problem: acceleration is the second derivative of position with respect to time. Answer: (B) $2.5\ \mathrm{ms^{-2}}$.

    Concept and Intuition

    Given any position–time relation $s(t)$, velocity is $v = \dfrac{ds}{dt}$ and acceleration is $a = \dfrac{dv}{dt}$. Since the given relation is quadratic in $t$, the motion is uniformly accelerated (constant acceleration), so differentiating twice gives a time-independent constant.

    Step-by-Step Solution

    1. Rewrite $4s = 5t^{2}+8t$ as $s = \dfrac{5}{4}t^{2} + 2t$.
    2. Differentiate once for velocity: $v = \dfrac{ds}{dt} = \dfrac{5}{2}t + 2$.
    3. Differentiate again for acceleration: $a = \dfrac{dv}{dt} = \dfrac{5}{2} = 2.5\ \mathrm{ms^{-2}}$. …

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12. AP EAPCET 2022Set ap-2022-07-11-FN1 markMCQ
    Q.A particle is moving along straight line. If its location at time t is given by $x(t) = \alpha t e^{-t/\tau}$; $\alpha = 1\ \mathrm{m\,s^{-1}}$, $\tau = 1$ s. The velocity of the particle at t = 2 s is (A) $\dfrac{-1}{e}\ \mathrm{m\,s^{-1}}$ (B) $\dfrac{-1}{e^2}\ \mathrm{m\,s^{-1}}$ (C) $\dfrac{-2}{e}\ \mathrm{m\,s^{-1}}$ (D) $\dfrac{-2}{e^2}\ \mathrm{m\,s^{-1}}$

    ›Reveal solutionSolution

    Velocity is $v = dx/dt$. Differentiating the given product-of-exponential expression and plugging in $t = 2$ s gives $v = -1/e^2\ \mathrm{m\,s^{-1}}$.

    Concept and Intuition

    Whenever position is given as an explicit function of time, velocity is simply its time derivative. Here $x(t) = \alpha t e^{-t/\tau}$ is a product of $t$ and a decaying exponential, so the product rule is required.

    Step-by-Step Solution

    1. Write $x(t) = \alpha t\, e^{-t/\tau}$.
    2. Differentiate using the product rule:

    $v(t) = \frac{dx}{dt} = \alpha e^{-t/\tau} + \alpha t \left(-\frac{1}{\tau}\right) e^{-t/\tau} = \alpha e^{-t/\tau}\left(1 - \frac{t}{\tau}\right)$

    3. Substitute $\alpha = 1\ \mathrm{m\,s^{-1}}$, $\tau = 1\ \mathrm{s}$, $t = 2\ \mathrm{s}$: …

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