A body of unit mass ($1\ \text{kg}$) moves in a plane, described by two velocity–time graphs. The $x$-velocity graph is triangular: $v_x$ rises linearly from $0$ at $t=0$ to $2\ \text{m s}^{-1}$ at $t=1\ \text{s}$, then falls linearly back to $0$ at $t=2\ \text{s}$, and remains $0$ for $t>2\ \text{s}$. The $y$-velocity graph rises linearly from $0$ at $t=0$ to $1\ \text{m s}^{-1}$ at $t=1\ \text{s}$, and then stays constant at $1\ \text{m s}^{-1}$ for $t>1\ \text{s}$. Find the force acting on the body as a function of time.

A body of unit mass (1kg) moves in a plane, described by two velocity–time graphs. The x-velocity graph is tri

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Concept understanding — Newton Second Law

Newton's Second Law: The Law That Connects Force and Motion

Imagine you're pushing a shopping cart. If you push gently, it moves slowly. Push harder, and it speeds up faster. Now imagine the cart is full of groceries — even with the same push, it accelerates much more slowly than an empty cart. This everyday experience is exactly what Newton's Second Law captures.

The Intuition First

Two things matter when you push something:

How hard you push — the force you apply.
How heavy the object is — its mass.

The harder you push, the more the object speeds up. The heavier the object, the less it speeds up for the same push. So acceleration depends on both force and mass — and in opposite ways.

Note

"Acceleration" here means any change in velocity — speeding up, slowing down, or changing direction. It's not just "going faster."

The Precise Statement

Newton's Second Law says:

The acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass. The acceleration is in the same direction as the net force.

In one equation:

\vec{a} = \frac{\vec{F}_{\text{net}}}{m}

Or more commonly:

\vec{F}_{\text{net}} = m \vec{a}

Where:

$\vec{F}_{\text{net}}$ is the net force (the vector sum of all forces acting on the object) — measured in newtons (N)
$m$ is the mass of the object — measured in kilograms (kg)
$\vec{a}$ is the acceleration — measured in metres per second squared ( $\text{m/s}^2$ )

\vec{F}_{\text{net}} = m \vec{a}

What This Really Means

Force causes acceleration, not velocity. A constant net force produces constant acceleration — meaning the velocity keeps changing at a steady rate. If you stop pushing, the net force becomes zero, and acceleration becomes zero (the object continues at constant velocity — that's Newton's First Law).

Mass is a measure of inertia. The more mass an object has, the harder it is to change its motion. A truck needs a much larger force than a bicycle to achieve the same acceleration.

Direction matters. Force and acceleration are vectors — they point the same way. If you push north, the acceleration is north. If multiple forces act, you must add them as vectors to find the net force.

A Simple Example

A 2 kg block is pushed with a net force of 10 N to the right.

a = \frac{F_{\text{net}}}{m} = \frac{10\ \text{N}}{2\ \text{kg}} = 5\ \text{m/s}^2

The block accelerates at $5\ \text{m/s}^2$ to the right. Every second, its velocity increases by 5 m/s in that direction. …

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Acceleration is the slope of a velocity–time graph, and with $m=1\ \text{kg}$ the force numerically equals the acceleration. Reading the slopes piecewise gives a force that changes in three time intervals.

Concept

$\mathbf F=m\mathbf a$ with $m=1\ \text{kg}$ , and $a_x=\dfrac{dv_x}{dt}$ , $a_y=\dfrac{dv_y}{dt}$ are the slopes of the two graphs.

$x$ -component (triangular $v_x$ )

$0<t<1\ \text{s}$ : $v_x$ goes $0\to2$ , slope $a_x=+2\ \text{m s}^{-2}\Rightarrow F_x=+2\ \text{N}$ .
$1<t<2\ \text{s}$ : $v_x$ goes $2\to0$ , slope $a_x=-2\ \text{m s}^{-2}\Rightarrow F_x=-2\ \text{N}$ .
$t>2\ \text{s}$ : $v_x=0$ , so $F_x=0$ .

$y$ -component

$0<t<1\ \text{s}$ : $v_y$ goes $0\to1$ , slope $a_y=+1\ \text{m s}^{-2}\Rightarrow F_y=+1\ \text{N}$ .
$t>1\ \text{s}$ : $v_y=1\ \text{m s}^{-1}$ constant, so $F_y=0$ .

Force as a function of time …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

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KEAM 2026Set eng-2026-04174 marksMCQ
$p = (3t^2 + 2t + 1)\ \text{kg m s}^{-1}$
›Reveal solutionSolution

Force is the time-derivative of momentum.

Given $p=3t^2+2t+1$ ,
$F=\frac{dp}{dt}=6t+2.$
At $t=3\ \text{s}$ : …
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KEAM 2026Set eng-2026-04194 marksMCQ
$g$
›Reveal solutionSolution

Force equals momentum carried away per second: $n\,m\,v$ .

Each bullet carries momentum $mv=(0.01\ \text{kg})(20\ \text{m s}^{-1})=0.2\ \text{kg m s}^{-1}$ . …
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KEAM 2026Set eng-2026-04204 marksMCQ
$m$
›Reveal solutionSolution

Use Newton's second law $F=m\dfrac{dv}{dt}$ with $F=kt$ and integrate from rest.

Newton's second law gives $m\dfrac{dv}{dt}=kt$ . Separating variables, $dv=\dfrac{kt}{m}\,dt$ . Integrating from $0$ to $t$ with $v(0)=0$ :

$v=\dfrac{k}{m}\cdot\dfrac{t^2}{2}=\dfrac{kt^2}{2m}.$ …
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KEAM 2026Set pha-2026-0419F4 marksMCQ
$ms^{-2}$
›Reveal solutionSolution

Resultant of perpendicular forces = $\sqrt{3^2 + 4^2} = 5$ N; $m = F/a = 5/2 = 2.5$ kg.

Two perpendicular forces combine as $F = \sqrt{3^2 + 4^2} = 5$ N. With acceleration $a = 2$ m s⁻², mass …
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KEAM 2025Set eng-2025-04264 marksMCQ
$25\ ms^{-1}$
›Reveal solutionSolution

Force equals mass times acceleration, and acceleration is the change in velocity over time: $F = m\,\Delta v/t = 1\ \text{N}$ .

Given: $m = 200\ \text{g} = 0.2\ \text{kg}$ , $\Delta v = 25\ \text{ms}^{-1}$ , $t = 5\ \text{s}$ .

Acceleration: …
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KEAM 2025Set eng-2025-04284 marksMCQ
$10 \text{ ms}^{-1}$
›Reveal solutionSolution

$a=F/m=25\ \text{ms}^{-2}$ ; $v=u+at=10+25\times10=260\ \text{ms}^{-1}$ .

The acceleration produced by the force is

$a=\frac{F}{m}=\frac{50}{2}=25\ \text{ms}^{-2}.$ …
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KEAM 2025Set eng-2025-04294 marksMCQ
$300\,\text{ms}^{-1}$
›Reveal solutionSolution

Force $=$ rate of momentum ejection $=\dfrac{(0.04)(300)\times25}{60}=5\,\text{N}$ .

Momentum carried per bullet. Mass $m=40\,\text{g}=0.04\,\text{kg}$ , speed $v=300\,\text{m/s}$ :
$p = mv = 0.04\times300 = 12\\,\text{kg·m/s}.$
Rate of firing. $25$ bullets per minute $= \dfrac{25}{60}\,\text{s}^{-1}$ . …
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KEAM 2025Set eng-2025-04294 marksMCQ
$\vec{F} = \hat{i} + 2\hat{j} - 2\hat{k}$
›Reveal solutionSolution

$|\vec{F}|=3\,\text{N}$ , $a=2\,\text{m/s}^2$ , so $m=F/a=1.5\,\text{kg}$ .

Magnitude of the force. For $\vec{F}=\hat{i}+2\hat{j}-2\hat{k}$ ,

$|\vec{F}| = \sqrt{1^2+2^2+(-2)^2} = \sqrt{1+4+4} = \sqrt{9} = 3\\,\text{N}.$ …
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KEAM 2025Set pha-2025-0424F4 marksMCQ
$F_1, F_2$
›Reveal solutionSolution

In equilibrium the three forces sum to zero, so $\vec F_1+\vec F_2=-\vec F_3$ . Take away $F_3$ and the leftover resultant has magnitude $F_3$ ; acceleration $=F_3/m$ .

Derivation: Equilibrium requires
$\vec F_1+\vec F_2+\vec F_3 = 0 \;\Rightarrow\; \vec F_1+\vec F_2 = -\vec F_3.$
When $F_3$ is removed, the net force is

$\vec F_{net} = \vec F_1+\vec F_2 = -\vec F_3,$ …
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KEAM 2025Set pha-2025-0424A4 marksMCQ
$Q. If the forces acting on two bodies of masses 2 kg and 3 kg are same , then the ratio of their respective accelerations is (A) 1 : 1 (B) 1 : 2 (C) 2 : 3 (D) 3 : 2 (E) 4 : 9$
›Reveal solutionSolution

From $F=ma$ with equal $F$ , $a \propto 1/m$ . So $a_1:a_2 = \dfrac{1}{2}:\dfrac{1}{3} = 3:2$ . …

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KEAM 2025Set pha-2025-0429F4 marksMCQ
$x = 3t + 7t^2$
›Reveal solutionSolution

Differentiating $x = 3t + 7t^2$ twice gives $a = 14\,ms^{-2}$ ; with $m = 0.2$ kg, $F = ma = 2.8\,\text{N}$ .

The displacement is $x = 3t + 7t^2$ . Velocity and acceleration are

$v = \frac{dx}{dt} = 3 + 14t, \qquad a = \frac{dv}{dt} = 14\,ms^{-2}.$ …
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KEAM 2024Set eng-2024-06054 marksMCQ
$4\,\mathrm{ms}^{-2}$
›Reveal solutionSolution

Buoyancy is fixed at $0.84$ N; to descend at $4\ \mathrm{ms^{-2}}$ the total mass must be 140 g, so add 80 g.

Step 1: Moving up with acceleration $a=4\ \mathrm{ms^{-2}}$ : $B-mg=ma\Rightarrow B=m(g+a)=0.060\times14=0.84$ N.

Step 2: To descend with the same acceleration, weight now exceeds buoyancy: $Mg-B=Ma\Rightarrow B=M(g-a)=M\times6$ . …
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