Q.A body moves in a straight line (unidirectionally) under a source that delivers energy at a constant power. Which of the following displacement–time ($d$ versus $t$) curves correctly represents its motion?

Showing the 12 most recent of 17 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.A 1000 kg car moves at $20\ ms^{-1}$ power required to maintain speed against 500 N resistive force is (A) 5 kW (B) 10 kW (C) 20 kW (D) 50 kW

   ›Reveal solutionSolution

   Power to maintain constant speed against a resistive force is simply $P = Fv$; here it's 10 kW.

   Concept and Intuition

   When a car moves at constant velocity, its net force is zero, so the driving force from the engine must exactly balance the resistive force. Power delivered by any force is the dot product of force and velocity, which for force and velocity in the same direction is just their product.

   Step-by-Step Solution

   1. At constant speed, net force = 0, so engine's driving force $F = 500\ N$ (equal and opposite to resistance).
   2. Power required: $P = F \times v = 500 \times 20 = 10{,}000\ W$.
   3. Convert to kW: $10{,}000\ W = 10\ kW$. …

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2. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.If force, $F = kv^n$ ($k$ is constant) and power delivered is independent of velocity 'v', then $n$ equals to (A) 0 (B) 1 (C) $-1$ (D) $-2$

   ›Reveal solutionSolution

   Power is force times velocity; substituting $F=kv^n$ shows $P\propto v^{n+1}$, which is independent of $v$ only when $n=-1$.

   Concept and Intuition

   Power delivered by a force moving at velocity $v$ is $P = Fv$. If the force itself depends on velocity as $F = kv^n$, then the power becomes $P = k v^{n+1}$. For $P$ to have no dependence on $v$ at all, the net power of $v$ appearing in the expression must be zero.

   Step-by-Step Solution

   1. Write power: $P = F\cdot v$.
   2. Substitute $F = kv^n$: $P = k v^n \cdot v = k v^{n+1}$. …

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3. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.A body is moving in a straight line with constant power. The distance moved by the body in time $t$ is proportional to (A) $\sqrt{t}$ (B) $t^2$ (C) $t^{3/2}$ (D) $t^{3/4}$

   ›Reveal solutionSolution

   With constant power, $v\propto\sqrt t$ and integrating velocity over time gives distance $\propto t^{3/2}$.

   Concept and Intuition

   Power is the rate of doing work, $P=Fv$. Since $F=m\,dv/dt$, we get a differential equation relating $v$ and $t$ for constant $P$. Solving it shows the speed grows as $\sqrt{t}$, and integrating speed over time gives how distance grows.

   Step-by-Step Solution

   1. $P = Fv = m\dfrac{dv}{dt}\,v$, constant.
   2. Rearranging: $v\,dv = \dfrac{P}{m}\,dt$.
   3. Integrating from $0$ to $t$ (starting from rest): $\dfrac{v^2}{2}=\dfrac{P}{m}t \Rightarrow v=\sqrt{\dfrac{2Pt}{m}} \propto t^{1/2}$. …

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4. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.A body is moving along a straight line under the influence of a constant power source. If the relation between the displacement (s) of the body and time (t) is $s \propto t^x$, then $x =$ (A) $1$ (B) $2$ (C) $\dfrac{2}{3}$ (D) $\dfrac{3}{2}$

   ›Reveal solutionSolution

   Under constant power, velocity grows as $t^{1/2}$ and displacement as $t^{3/2}$, so $x=3/2$.

   Concept and Intuition

   Constant power means $P=Fv=mv\frac{dv}{dt}$ is constant, which is a differential equation directly relating $v$ and $t$ (not $F$ and $t$ as in constant force). Solving it shows velocity itself grows with time as a power law, and integrating once more gives displacement's power law.

   Step-by-Step Solution

   1. Constant power: $P=Fv=m\dfrac{dv}{dt}\cdot v=$ constant $=k$ (say).
   2. Rewrite: $v\,dv=\dfrac{k}{m}\,dt$. Integrate both sides (starting from rest, $v=0$ at $t=0$): $\frac{v^2}{2}=\frac{k}{m}t \;\Rightarrow\; v^2\propto t \;\Rightarrow\; v\propto t^{1/2}.$ …

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5. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.A car of mass 2000 kg is accelerating from rest. If its engine is supplying constant power of 10 kW, then the velocity of the car at a time of 10 s is (A) $15\ \mathrm{m\,s^{-1}}$ (B) $20\ \mathrm{m\,s^{-1}}$ (C) $5\ \mathrm{m\,s^{-1}}$ (D) $10\ \mathrm{m\,s^{-1}}$

   ›Reveal solutionSolution

   This tests the work-energy theorem for constant-power motion starting from rest, avoiding the trap of assuming constant acceleration. Answer: (D).

   Concept and Intuition

   When power (not force) is constant, the force $F=P/v$ decreases as speed increases, so acceleration is not constant — the usual $v=u+at$ does not directly apply. Instead, use energy: at constant power, the work done equals $P\times t$, and this must equal the gain in kinetic energy.

   Step-by-Step Solution

   1. Work done by the engine in time $t$ at constant power $P$: $W = Pt$.
   2. Since the car starts from rest, all this work becomes kinetic energy: $Pt = \dfrac12 mv^2 - 0$.
   3. Solve for $v$: $v = \sqrt{\dfrac{2Pt}{m}}$.
   4. Substitute $P=10\,000\ \text{W}$, $t=10\ \text{s}$, $m=2000\ \text{kg}$: …

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6. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.A body of mass 'M' is moving with a uniform speed of 'v' on a frictionless horizontal surface under the influence of two forces $F_1$ and $F_2$ as shown in the figure. The net power of the system is [FIGURE] (a block of mass M on a horizontal surface, with force $F_1$ acting rightward on the left side and force $F_2$ acting leftward on the right side, both forces pointing toward the block) (A) $(F_1 - F_2)v$ (B) $0.5(F_1 + F_2)v$ (C) $(F_1 + F_2)v$ (D) zero

   ›Reveal solutionSolution

   This tests the work-energy theorem: for a body moving at truly constant (uniform) speed, the net power delivered by all forces is always zero — regardless of the individual force values. Answer: (D).

   Concept and Intuition

   Power is the rate of doing work, and kinetic energy is directly tied to total work done (work-energy theorem). If the speed genuinely stays uniform, kinetic energy is constant in time, so the net work done — and therefore the net power delivered by all forces combined — must be zero at every instant. This is true even though two individual forces are acting on the body.

   Step-by-Step Solution

   1. The body moves on a frictionless horizontal surface with uniform speed $v$, under only two horizontal forces $F_1$ and $F_2$ acting toward each other (opposing directions) along the line of motion.
   2. For the speed to remain truly uniform (not just momentarily equal but constant throughout), the net force on the body must be zero at all times — otherwise it would accelerate and speed would change. This forces $F_1=F_2$.
   3. Kinetic energy $=\tfrac12 Mv^2$ is therefore constant (since $v$ doesn't change).
   4. By the work-energy theorem, the total work done by all forces on the body over any time interval is $\Delta KE = 0$; correspondingly, the net power (rate of total work) delivered to the body is zero at every instant. …

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7. AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
   Q.A substance of mass $m$ requires a power input of $P$ to remain in the molten state at its melting point. When the power is turned off, the sample completely solidifies in time $t$. The latent heat of the substance is (A) $\dfrac{Pt}{m}$ (B) $\dfrac{Pm}{t}$ (C) $\dfrac{m}{Pt}$ (D) $\dfrac{t}{Pm}$

   ›Reveal solutionSolution

   The power $P$ that was needed to keep the substance molten equals its steady rate of heat loss to the surroundings; that same rate removes the latent heat when the power is cut. Answer: (A) $\dfrac{Pt}{m}$.

   Concept and Intuition

   At the melting point, a pure substance can sit as liquid (or partly liquid/partly solid) at constant temperature — any heat you add there does NOT raise its temperature, it goes into changing phase (or, run in reverse, holding it in the liquid phase against heat losses). If the substance's temperature is not changing while power $P$ is supplied, that power must be exactly compensating the rate of heat loss to the environment (otherwise it would slowly freeze even with the heater on). So the environment is steadily draining heat from the sample at rate $P$, regardless of whether the heater is on or off — the heater's job was simply to replace that loss.

   Step-by-Step Solution

   1. Before power-off: to remain molten at constant temperature (melting point), heat supplied = heat lost to surroundings, both equal to $P$ (power). This tells us the ambient heat-loss rate for this sample is $P$.
   2. Power off: the heater no longer replaces the loss, so the sample now loses heat to the surroundings at that same rate $P$ (same temperature difference, same conditions, so same rate).
   3. This heat loss is entirely what freezes the sample (heat leaving = latent heat given up, since temperature stays at the melting point throughout the freezing process). …

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8. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.A particle of mass 'm' at rest on a rough horizontal surface with a coefficient of friction 'μ' is given a velocity 'u'. The average power imparted by friction before it stops (A) Zero (B) $\frac{1}{2}\mu mgu$ (C) $\mu mgu$ (D) $2\mu mgu$

   ›Reveal solutionSolution

   Average power is total work done by friction (equal to the initial kinetic energy) divided by the total time taken to stop, giving $\tfrac12\mu m g u$.

   Concept and Intuition

   "Average power" here means total energy dissipated divided by total time — not the instantaneous power at any moment (which actually starts at $\mu mgu$ and falls linearly to zero as the particle slows). The whole kinetic energy $\tfrac12mu^2$ is dissipated by friction, and this happens over the stopping time $t=u/(\mu g)$.

   Step-by-Step Solution

   1. Friction force $f=\mu mg$, deceleration $a=\mu g$.
   2. Time to stop from speed $u$: $t=\dfrac{u}{\mu g}$.
   3. Work done by friction (magnitude) $=$ initial KE $=\dfrac12 mu^2$. …

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9. AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
   Q.The charge flowing through a resistance 'R' varies with time as $q = at - bt^2$ where a,b are positive constants. The total heat produced in R is (A) $\dfrac{a^3 R}{6b}$ (B) $\dfrac{a^3 R}{2b}$ (C) $\dfrac{a^3 R}{3b}$ (D) $\dfrac{a^3 R}{b}$

   ›Reveal solutionSolution

   This tests using $i=dq/dt$ and then integrating $i^2R$ over time to find total heat — the answer is $\dfrac{a^3R}{6b}$.

   Concept and Intuition

   Heat produced in a resistor over an interval is $H=\int i^2R\,dt$ (Joule heating), not simply $q^2R$ or similar shortcuts — because current itself varies with time here, so we must integrate its square. The first step is always to get the actual instantaneous current from the given charge function.

   Step-by-Step Solution

   1. Charge: $q(t) = at - bt^2$. Current: $i = \dfrac{dq}{dt} = a - 2bt$.
   2. This current starts at $a$ (at $t=0$) and decreases linearly, becoming zero at $t = \dfrac{a}{2b}$ — beyond this instant the current (as modeled) would go negative, i.e., the physical flow described has effectively ended. So the relevant interval for a single "pulse" of current is $t\in\left[0, \dfrac{a}{2b}\right]$.
   3. Heat generated: $H = \displaystyle\int_0^{a/2b} i^2 R\,dt = R\int_0^{a/2b} (a-2bt)^2\,dt$.
   4. Substitute $x = a-2bt \Rightarrow dx = -2b\,dt \Rightarrow dt = -\dfrac{dx}{2b}$. Limits: $t=0\Rightarrow x=a$; $t=\frac{a}{2b}\Rightarrow x=0$. …

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10. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
    Q.A toy of mass 20 g at rest acquires a velocity $(3\hat{i}-2\hat{j})\text{ms}^{-1}$ in 2 seconds. Then the power of the toy is (A) 0.975 W (B) 0.325 W (C) 1.3 W (D) 0.065 W

    ›Reveal solutionSolution

    This tests the work-energy theorem: power delivered equals the kinetic energy gained

    divided by time. Answer: 0.065 W.

    Concept and Intuition

    Power is the rate of doing work. If a body starts from rest and ends up with some

    kinetic energy after a time $t$, all of that kinetic energy must have come from the

    work done by the driving force, so the average power is just that kinetic energy

    divided by the time taken — we don't need the actual force or the path.

    Step-by-Step Solution

    1. Mass $m = 20\text{ g} = 0.02$ kg. Final velocity $\vec v = (3\hat i - 2\hat j)\text{ ms}^{-1}$.
    2. Speed squared: $v^2 = 3^2+(-2)^2 = 9+4 = 13\text{ m}^2\text{s}^{-2}$.
    3. Since the toy starts at rest, kinetic energy gained: $KE=\tfrac12 mv^2=\tfrac12(0.02)(13)=0.13\text{ J}$ …

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11. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
    Q.A bead of mass $400$ g is moving along a straight line under a force that delivers a constant power $1.2$ W to the bead. If the bead is initially at rest, the speed it attains after $6$ sec in $\mathrm{ms^{-1}}$ (A) $5$ (B) $4$ (C) $6$ (D) $3$

    ›Reveal solutionSolution

    Constant power delivered over time $t$ does work $Pt$; equating this to the kinetic energy gained from rest gives $v=6\ \mathrm{ms^{-1}}$.

    Concept and Intuition

    Power is the rate of doing work. If power is held constant (not force), the total work done over a time interval is simply $P\times t$, regardless of how velocity varies during that interval. By the work-energy theorem, this work equals the change in kinetic energy.

    Step-by-Step Solution

    1. Work done in $t=6$ s at constant power $P=1.2$ W: $W = Pt = 1.2\times6 = 7.2$ J.
    2. Since the bead starts from rest, this work equals the final kinetic energy: $W=\dfrac12mv^2$.
    3. $7.2 = \dfrac12(0.4)v^2 = 0.2v^2$. …

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12. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
    Q.A battery with a 12 V emf has an initial charge of 80 A.h. If the potential across the terminals stays constant until battery is completely discharged, then this battery can deliver energy at the rate of 120 W for a time (A) 16 h (B) 8 h (C) 4 h (D) 5 h

    ›Reveal solutionSolution

    Compute total stored energy from EMF and charge capacity, then divide by the power delivery rate.

    Concept and Intuition

    A battery's charge capacity in ampere-hours (A·h), multiplied by its (constant) terminal voltage, gives the total energy it can deliver in watt-hours. Dividing this total energy by a constant power output gives the time for which that power can be sustained.

    Step-by-Step Solution

    1. Total energy $= V \times Q = 12\,\text{V} \times 80\,\text{A·h} = 960\,\text{Wh}$. …

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