JEE Main Motion in a Straight Line Previous Year Questions (PYQs) MCQs are extremely important for understanding the concepts of kinematics and improving problem-solving skills for competitive examinations. These questions cover important topics such as displacement, velocity, acceleration, equations of motion, graphical analysis, relative motion, and motion under gravity. Practising PYQs helps students understand the exam pattern, difficulty level, and frequently asked concepts while strengthening their conceptual clarity and numerical solving techniques.
JEE Main MCQs, PYQs, Previous Year Questions with Detailed Solutions for Motion in a Straight Line Class 11 Physics
Practice JEE Main Motion in a Straight Line MCQs, PYQs, and previous year questions with detailed solutions for Class 11 Physics. These important questions help students strengthen concepts of displacement, velocity, acceleration, equations of motion, and graphical analysis while improving numerical problem-solving skills for JEE Main preparation.
“Students should also study Equations of Motion Under Gravity for free fall, upward and vertical downward motion“
Why should students practice JEE Main Motion in a Straight Line PYQs?
Practising previous year questions helps students understand the exam pattern, important concepts, and the level of difficulty asked in JEE Main. It also improves numerical problem-solving skills and strengthens conceptual understanding of kinematics.
“Read more about Derive Equations of Motion Using Calculus Method, Solved Numerical Examples“
Which topics are covered in Motion in a Straight Line MCQs and PYQs?
These questions generally cover displacement, distance, speed, velocity, acceleration, equations of motion, relative motion, graphical analysis, and motion under gravity. They are important for building a strong foundation in Class 11 Physics.
“Build strong concepts by studying Derive Equations of Uniformly Accelerated Motion“
JEE Main Question
If the velocity of a body related to displacement $x$ is given by $v = \sqrt{5000 + 24x}\text{ m/s}$, then the acceleration of the body is _______ $\text{m/s}^2$.
(JEE Main 2021, 27 Aug Shift-I)
Solution :
Answer : 12
Velocity as a function of displacement : $$v = \sqrt{5000 + 24x} $$
When velocity is given as a function of position ($x$) rather than time ($t$), acceleration ($a$) is calculated using the chain rule :
$$a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}$$
Since $\dfrac{dx}{dt} = v$, the expression becomes :
$$a = v\frac{dv}{dx} $$
$$\frac{dv}{dx} = \frac{d}{dx}\left(\sqrt{5000 + 24x}\right)$$
$$\frac{dv}{dx} = \frac{1}{2\sqrt{5000 + 24x}} \cdot \frac{d}{dx}(5000 + 24x)$$
$$\frac{dv}{dx} = \frac{1}{2\sqrt{5000 + 24x}} \cdot (24)$$
$$\frac{dv}{dx} = \frac{12}{\sqrt{5000 + 24x}}$$
$$a = v\frac{dv}{dx}$$
$$a = \left(\sqrt{5000 + 24x}\right) \times \left(\frac{12}{\sqrt{5000 + 24x}}\right)$$
$$a = 12\text{ m/s}^2$$
The acceleration of the body is a constant $12\text{ m/s}^2$.
“For complete preparation, also study JEE Units and Measurements PYQs Previous Year Questions and Solutions“
JEE Main Question
The relation between time $t$ and distance $x$ for a moving body is given as $t = mx^2 + nx$, where $m$ and $n$ are constants. The retardation of the motion is (where $v$ stands for velocity) :
(A) $2mv^3$
(B) $2mnv^3$
(C) $2nv^3$
(D) $2n^2v^3$
(JEE Main 2021, 25 July Shift-II)
Solution :
Answer: (A)
Time-displacement relation : $$t = mx^2 + nx $$
Differentiate with respect to time ($t$) on both sides :
$$\frac{dt}{dt} = \frac{d}{dt}(mx^2 + nx)$$
$$1 = 2mx\frac{dx}{dt} + n\frac{dx}{dt}$$
Since velocity is defined as the rate of change of displacement ($v = \dfrac{dx}{dt}$), substitute $v$ into the equation :
$$1 = 2mxv + nv$$
$$1 = v(2mx + n)$$
$$v = \frac{1}{2mx + n} $$
To find acceleration ($a = \dfrac{dv}{dt}$), we differentiate the velocity equation with respect to time ($t$). It is easier to differentiate the implicit form $v^{-1} = 2mx + n$:
$$\frac{d}{dt}(2mx + n) = \frac{d}{dt}\left(\frac{1}{v}\right)$$
Differentiating both sides with respect to $t$:
$$2m\frac{dx}{dt} = -\frac{1}{v^2}\frac{dv}{dt}$$
Substitute $\dfrac{dx}{dt} = v$ and $\dfrac{dv}{dt} = a$:
$$2mv = -\frac{1}{v^2}a$$
$$a = -2mv^3$$
Retardation is defined simply as negative acceleration (the magnitude of deceleration) :
$$\text{Retardation} = -a = -(-2mv^3) = 2mv^3$$
The retardation of the moving body is $2mv^3$.
Hence, the correct option is (A).
For complete preparation, also study NCERT Exemplar Solutions Motion In a Straight Line Class 11 Physics Chapter 2
JEE Main Question
The instantaneous velocity of a particle moving in a straight line is given as $v = at + bt^2$, where $a$ and $b$ are constants. The distance travelled by the particle between $1\text{ s}$ and $2\text{ s}$ is :
(A) $3a + 7b$
(B) $\frac{3}{2}a + \frac{7}{3}b$
(C) $\frac{a}{2} + \frac{b}{3}$
(D) $\frac{3}{2}a + \frac{7}{2}b$
(JEE Main 2021, 25 July Shift-II)
Solution :
Answer: (B)
$$v = \frac{ds}{dt}$$
$$ds = v \, dt$$
Substitute the given expression for $v$ :
$$ds = (at + bt^2) dt$$
To find the total distance ($S$) covered between $1\text{ s}$ and $2\text{ s}$, integrate both sides of the equation :
$$S = \int_{1}^{2} (at + bt^2) dt$$
$$S = \left[ a\frac{t^2}{2} + b\frac{t^3}{3} \right]_{1}^{2}$$
$$S = \left( a\frac{(2)^2}{2} + b\frac{(2)^3}{3} \right) – \left( a\frac{(1)^2}{2} + b\frac{(1)^3}{3} \right)$$
$$S = \left( \frac{4a}{2} + \frac{8b}{3} \right) – \left( \frac{a}{2} + \frac{b}{3} \right)$$
$$S = \left( \frac{4a}{2} – \frac{a}{2} \right) + \left( \frac{8b}{3} – \frac{b}{3} \right)$$
$$S = \frac{3a}{2} + \frac{7b}{3}$$
The total distance travelled by the particle between $1\text{ s}$ and $2\text{ s}$ is $\frac{3}{2}a + \frac{7}{3}b$.
Hence, the correct option is (B).
To strengthen your concepts, learn about Relative Velocity Formula Explained With Solved Examples
JEE Main Question
A boy reaches the airport and finds that the escalator is not working. He walks up the stationary escalator in time $t_1$. If he remains stationary on a moving escalator, then the escalator takes him up in time $t_2$. The time taken by him to walk up on the moving escalator will be :
(A) $\frac{t_1t_2}{t_2 – t_1}$
(B) $\frac{t_1 + t_2}{2}$
(C) $\frac{t_1t_2}{t_1 + t_2}$
(D) $t_2 – t_1$
(JEE Main 2021, 20 July Shift-II)
Solution
Answer: (C)
Let $L$ be the total length of the escalator.
Velocity of the boy ($v_1$) when walking up a stationary escalator : $$v_1 = \frac{\text{Distance}}{\text{Time}} = \frac{L}{t_1} $$
Velocity of the escalator ($v_2$) when the boy stands still on it : $$v_2 = \frac{\text{Distance}}{\text{Time}} = \frac{L}{t_2} $$
When the boy walks up a moving escalator, both velocities act in the same upward direction. Therefore, his net velocity ($v_{\text{net}}$) relative to the ground is the sum of both individual velocities :
$$v_{\text{net}} = v_1 + v_2$$
$$v_{\text{net}} = \frac{L}{t_1} + \frac{L}{t_2}$$
$$v_{\text{net}} = L \left( \frac{1}{t_1} + \frac{1}{t_2} \right) = L \left( \frac{t_1 + t_2}{t_1t_2} \right) $$
The total time $t$ taken to cover the same distance $L$ at this combined net velocity is :
$$t = \frac{\text{Distance}}{v_{\text{net}}} = \frac{L}{v_{\text{net}}}$$
$$t = \frac{L}{L \left( \dfrac{t_1 + t_2}{t_1t_2} \right)}$$
$$t = \frac{1}{\left( \dfrac{t_1 + t_2}{t_1t_2} \right)}$$
$$t = \frac{t_1t_2}{t_1 + t_2}$$
The time taken by the boy to walk up on the moving escalator is $\dfrac{t_1t_2}{t_1 + t_2}$.
Hence, the correct option is (C).
JEE Main Question
The velocity of a particle is given by $v = v_0 + gt + Ft^2$. If its position is $x = 0$ at $t = 0$, then its displacement after time $t = 1\text{ s}$ is :
(A) $v_0 + g + F$
(B) $v_0 + \frac{g}{2} + \frac{F}{3}$
(C) $v_0 + \frac{g}{2} + F$
(D) $v_0 + 2g + 3F$
(JEE Main 2021, 17 March Shift-II)
Solution
Answer : (B)
$$v = \frac{dx}{dt}$$
$$dx = v \, dt$$
Substitute the given function of $v$ :
$$dx = (v_0 + gt + Ft^2) dt$$
To find the final displacement at $t = 1\text{ s}$, integrate both sides of the equation from the initial limits ($t = 0, x = 0$) to the final limits ($t = 1, x = x$) :
$$\int_{0}^{x} dx = \int_{0}^{1} (v_0 + gt + Ft^2) dt$$
$$[x]_{0}^{x} = \left[ v_0t + g\frac{t^2}{2} + F\frac{t^3}{3} \right]_{0}^{1}$$
$$x – 0 = \left( v_0(1) + \frac{g(1)^2}{2} + \frac{F(1)^3}{3} \right) – (0)$$
$$x = v_0 + \frac{g}{2} + \frac{F}{3}$$
The total displacement of the particle after $t = 1\text{ s}$ is $v_0 + \dfrac{g}{2} + \dfrac{F}{3}$.
Hence, the correct option is (B).
JEE Main Question
A balloon is moving up in the air vertically above a point $A$ on the ground. When it is at a height $h_1$, a girl standing at a distance $d$ (point $B$) from $A$ sees it at an angle of $45^\circ$ with respect to the vertical. When the balloon climbs up a further height $h_2$, it is seen at an angle of $60^\circ$ with respect to the vertical if the girl moves further by a distance $2.464d$ (point $C$). Then, the height $h_2$ is : (Given, $\tan 30^\circ = 0.5774$)
(A) $1.464d$
(B) $d$
(C) $0.464d$
(D) $0.732d$
![JEE Main PYQ [2020, 5 Sep Shift-I] Solved Problem](https://physicsanandclasses.co.in/wp-content/uploads/2026/05/JEE-Main-2020-5-Sep-Shift-I-Solved-Problem.webp "JEE Main Motion in a Straight Line MCQs PYQs Previous Year Questions and Solutions 2")
(JEE Main 2020, 5 Sep Shift-I)
Solution
Answer : (B)
$$\tan 45^\circ = \frac{AD}{AB}$$
$$\tan 45^\circ = \frac{h_1}{d}$$
Since $\tan 45^\circ = 1$:
$$1 = \frac{h_1}{d} \implies h_1 = d
In the larger right-angled triangle $\Delta ACE$ :
$$\tan 30^\circ = \frac{AE}{AC}$$
$$\tan 30^\circ = \frac{h_1 + h_2}{3.464d}$$
Substitute the given value $\tan 30^\circ = 0.5774$ into the expression :
$$0.5774 = \frac{h_1 + h_2}{3.464d}$$
$$h_1 + h_2 = 0.5774 \times 3.464d$$
$$h_1 + h_2 \approx 2.0001d \approx 2d $$
Now, substitute $h_1 = d$ :
$$d + h_2 = 2d$$
$$h_2 = 2d – d$$
$$h_2 = d$$
The further height climbed by the balloon, $h_2$, is equal to $d$.
Hence, the correct option is (B).
JEE Main Question
The distance $x$ covered by a particle in one-dimensional motion varies with time $t$ as $x^2 = at^2 + 2bt + c$. If the acceleration of the particle depends on $x$ as $x^{-n}$, where $n$ is an integer, the value of $n$ is _______.
(JEE Main 2020, 9 Jan Shift-I)
Solution
Answer : 3
Given Equation is : $$x^2 = at^2 + 2bt + c$$
Differentiate both sides with respect to time ($t$) :
$$\frac{d}{dt}(x^2) = \frac{d}{dt}(at^2 + 2bt + c)$$
$$2x\frac{dx}{dt} = 2at + 2b$$
Since velocity $v = \dfrac{dx}{dt}$, substitute it into the equation and divide by 2 :
$$xv = at + b $$
$$\frac{d}{dt}(xv) = \frac{d}{dt}(at + b)$$
$$x\frac{dv}{dt} + v\frac{dx}{dt} = a$$
Since acceleration $A = \dfrac{dv}{dt}$ and velocity $v = \dfrac{dx}{dt}$ :
$$xA + v^2 = a$$
$$xA = a – v^2 $$
Since, $xv = at + b $, then $ v = \left(\dfrac{at + b}{x}\right)$, so the above equation becomes :
$$xA = a – \left(\frac{at + b}{x}\right)^2$$
$$xA = a – \frac{(at + b)^2}{x^2}$$
$$xA = \frac{ax^2 – (at + b)^2}{x^2}$$
Now substitute the value of $x^2$ from given equation in problem into the numerator :
$$xA = \frac{a(at^2 + 2bt + c) – (a^2t^2 + 2abt + b^2)}{x^2}$$
$$xA = \frac{\cancel{a^2t^2} + \cancel{2abt} + ac – \cancel{a^2t^2} – \cancel{2abt} – b^2}{x^2}$$
$$xA = \frac{ac – b^2}{x^2}$$
$$A = \frac{ac – b^2}{x^3}$$
Since $a$, $b$, and $c$ are constants, the term $(ac – b^2)$ is also a constant. Therefore, we can write the proportionality relation as :
$$A \propto \frac{1}{x^3}$$
$$A \propto x^{-3}$$
Comparing this directly with the given condition $A \propto x^{-n}$:
$$-n = -3 \implies n = 3$$
The value of the integer $n$ is 3.
JEE Main Question
A particle is moving with speed $v = b\sqrt{x}$ along the positive X-axis. Calculate the speed of the particle at time $t = \tau$ (assume that the particle is at the origin at $t = 0$).
(A) $\frac{b^2}{4}\tau$
(B) $\frac{b^2}{2}\tau$
(C) $b^2\tau$
(D) $\frac{b^2}{\sqrt{2}}\tau$
(JEE Main 2019, 12 April Shift-II)
Solution :
Answer: (B)
When velocity is a function of position ($x$), acceleration can be calculated using the formula :
$$a = v\frac{dv}{dx}$$
$$\frac{dv}{dx} = \frac{d}{dx}\left(b \cdot x^{1/2}\right) = b \cdot \left(\frac{1}{2}x^{-1/2}\right) = \frac{b}{2\sqrt{x}}$$
Now, substitute $v$ and $\dfrac{dv}{dx}$ into the acceleration formula :
$$a = \left(b\sqrt{x}\right) \times \left(\frac{b}{2\sqrt{x}}\right)$$
$$a = \frac{b^2}{2} $$
Since $b$ is a constant, the acceleration $a$ is also a constant.
We are given that at $t = 0$, the particle is at the origin ($x = 0$). Substituting $x = 0$ into given equation :
$$u = b\sqrt{0} = 0$$
Since the acceleration is uniform, we can directly apply the first equation of motion ($v = u + at$) :
$$v = 0 + \left(\frac{b^2}{2}\right)\tau$$
$$v = \frac{b^2}{2}\tau$$
Alternate Method : By Integration
We can also find the position as a function of time by rewriting the velocity definition ($v = \dfrac{dx}{dt}$):
$$\frac{dx}{dt} = b\sqrt{x}$$
Separate the variables to get the $x$ terms on one side and $t$ terms on the other:
$$\frac{dx}{\sqrt{x}} = b \, dt \implies x^{-1/2} \, dx = b \, dt$$
Integrate both sides using the given limits (from $x=0$ at $t=0$ to position $x$ at time $\tau$) :
$$\int_{0}^{x} x^{-1/2} \, dx = \int_{0}^{\tau} b \, dt$$
$$\left[ \frac{x^{1/2}}{1/2} \right]_{0}^{x} = [b \cdot t]_{0}^{\tau}$$
$$2\sqrt{x} = b\tau \implies \sqrt{x} = \frac{b\tau}{2}$$
Now substitute this value of $\sqrt{x}$ back into the original velocity equation ($v = b\sqrt{x}$):
$$v = b \left(\frac{b\tau}{2}\right)$$
$$v = \frac{b^2}{2}\tau$$
The speed of the particle at time $t = \tau$ is $\dfrac{b^2}{2}\tau$.
Hence, the correct option is (B).
JEE Main Question
The position of a particle as a function of time $t$, is given by :
$$x(t) = at + bt^2 – ct^3$$
where $a, b,$ and $c$ are constants. When the particle attains zero acceleration, then its velocity will be :
(A) $a + \frac{b^2}{2c}$
(B) $a + \frac{b^2}{4c}$
(C) $a + \frac{b^2}{3c}$
(D) $a + \frac{b^2}{c}$
(JEE Main 2019, 9 April Shift-II)
Solution
Answer: (C)
Velocity is the first derivative of position with respect to time :
$$v = \dfrac{dx}{dt}$$
$$v = \frac{d}{dt}(at + bt^2 – ct^3)$$
$$v = a(1) + b(2t) – c(3t^2)$$
$$v = a + 2bt – 3ct^2 $$
Acceleration is the derivative of velocity with respect to time :
$$A = \dfrac{dv}{dt}$$
$$A = \frac{d}{dt}(a + 2bt – 3ct^2)$$
$$A = 0 + 2b(1) – 3c(2t)$$
$$A = 2b – 6ct $$
Set the acceleration equation equal to zero ($A = 0$) :
$$2b – 6ct = 0$$
$$6ct = 2b$$
Solve for $t$ :
$$t = \frac{2b}{6c} = \frac{b}{3c} $$
Put the above value of t in equation : $v = a + 2bt – 3ct^2 $
$$v = a + 2b\left(\frac{b}{3c}\right) – 3c\left(\frac{b}{3c}\right)^2$$
$$v = a + \frac{2b^2}{3c} – 3c\left(\frac{b^2}{9c^2}\right)$$
$$v = a + \frac{2b^2}{3c} – \frac{b^2}{3c}$$
$$v = a + \frac{2b^2 – b^2}{3c}$$
$$v = a + \frac{b^2}{3c}$$
When the particle attains zero acceleration, its velocity is $a + \frac{b^2}{3c}$.
Hence, the correct option is (C).
Strengthen your fundamentals with JEE Main PYQs Previous Year Questions MCQs Topicwise Papers and Solutions Motion in a Straight Line
JEE Main Question
In a car race on a straight path, car $A$ takes a time $t$ less than car $B$ at the finish and passes the finishing point with a speed $v$ more than that of car $B$. Both the cars start from rest and travel with constant acceleration $a_1$ and $a_2$ respectively. Then $v$ is equal to :
(A) $\frac{2a_1a_2}{a_1 + a_2}t$
(B) $\sqrt{2a_1a_2}t$
(C) $\sqrt{a_1a_2}t$
(D) $\frac{a_1 + a_2}{2}t$
(JEE Main 2019, 9 Jan Shift-II)
Solution
Answer: (C)
Let the total distance of the race path to the finishing point be $S$. Both cars start from rest, meaning their initial velocities are zero ($u_A = u_B = 0$).
Let us define the times taken to complete the race :
Time taken by Car $A$ = $t_0$
Time taken by Car $B$ = $t_0 + t$ (since Car $A$ takes time $t$ less than Car $B$)
Using the second equation of motion ($S = ut + \frac{1}{2}at^2$) for both cars :
For Car $A$ :
$$S = \frac{1}{2}a_1t_0^2 $$
For Car $B$ :
$$S = \frac{1}{2}a_2(t_0 + t)^2 $$
Since both cars cover the exact same distance $S$, we can equate above equations :
$$\frac{1}{2}a_1t_0^2 = \frac{1}{2}a_2(t_0 + t)^2$$
$$\sqrt{a_1}t_0 = \sqrt{a_2}(t_0 + t)$$
$$\sqrt{a_1}t_0 = \sqrt{a_2}t_0 + \sqrt{a_2}t$$
$$t_0(\sqrt{a_1} – \sqrt{a_2}) = \sqrt{a_2}t$$
$$t_0 = \frac{\sqrt{a_2}t}{\sqrt{a_1} – \sqrt{a_2}} $$
Using the first equation of motion ($v = u + at$) for both cars :
Final velocity of Car $A$ : $v_A = a_1t_0$
Final velocity of Car $B$ : $v_B = a_2(t_0 + t)$
We are given that Car $A$ passes the finishing line with a speed $v$ more than Car $B$ :
$$v_A – v_B = v$$
$$v = a_1t_0 – a_2(t_0 + t)$$
$$v = (a_1 – a_2)t_0 – a_2t $$
Substitute the value of $t_0$ :
$$v = (a_1 – a_2)\left[\frac{\sqrt{a_2}t}{\sqrt{a_1} – \sqrt{a_2}}\right] – a_2t$$
$$v = \frac{(\sqrt{a_1} – \sqrt{a_2})(\sqrt{a_1} + \sqrt{a_2})\sqrt{a_2}t}{(\sqrt{a_1} – \sqrt{a_2})} – a_2t$$
$$v = (\sqrt{a_1} + \sqrt{a_2})\sqrt{a_2}t – a_2t$$
$$v = \sqrt{a_1}\sqrt{a_2}t + (\sqrt{a_2})^2t – a_2t$$
$$v = \sqrt{a_1a_2}t + \cancel{a_2t} – \cancel{a_2t}$$
$$v = \sqrt{a_1a_2}t$$
The extra relative velocity $v$ is equal to $\sqrt{a_1a_2}t$.
Hence, the correct option is (C).
“Strengthen your fundamentals with JEE PYQs Units and Measurements Previous Year Solved Questions“
JEE Main Question
An object moving with a speed of $6.25\text{ m/s}$ is decelerated at a rate given by $\dfrac{dv}{dt} = -2.5\sqrt{v}$, where $v$ is the instantaneous speed. The time taken by the object to come to rest would be :
(A) $2\text{ s}$
(B) $4\text{ s}$
(C) $8\text{ s}$
(D) $1\text{ s}$
(AIEEE 2011)
Solution
Answer: (A)
Initial speed at $t = 0$ ($u$) = $6.25\text{ m/s}$
Final speed ($v_f$) = $0\text{ m/s}$ (since it comes to rest)
Differential equation for deceleration :
$$\frac{dv}{dt} = -2.5\sqrt{v}$$
$$\frac{dv}{\sqrt{v}} = -2.5 \, dt$$
$$v^{-1/2} \, dv = -2.5 \, dt$$
Integrate both sides of the equation from the initial state ($t = 0, v = 6.25$) to the final rest state ($t = t, v = 0$) :
$$\int_{6.25}^{0} v^{-1/2} \, dv = \int_{0}^{t} -2.5 \, dt$$
$$\left[ \frac{v^{1/2}}{1/2} \right]_{6.25}^{0} = -2.5 [t]_{0}^{t}$$
$$\left[ 2\sqrt{v} \right]_{6.25}^{0} = -2.5(t – 0)$$
$$2\sqrt{0} – 2\sqrt{6.25} = -2.5t$$
$$0 – 2(2.5) = -2.5t$$
$$-5 = -2.5t$$
$$t = \frac{-5}{-2.5} = 2\text{ s}$$
The time taken by the object to come to rest is $2\text{ seconds}$.
Hence, the correct option is (A).
JEE Main Question
The velocity of a particle is given by $v = v_0 + gt + ft^2$. If its position is $x = 0$ at $t = 0$, then its displacement after unit time ($t = 1\text{ s}$) is :
(A) $v_0 + 2g + 3f$
(B) $v_0 + \frac{g}{2} + \frac{f}{3}$
(C) $v_0 + g + f$
(D) $v_0 + \frac{g}{2} + f$
(AIEEE 2007)
Solution
Answer: (B)
Velocity-time relationship : $$v = v_0 + gt + ft^2$$
Initial condition : At $t = 0$, position $x = 0$
Final condition : At $t = 1\text{ s}$ (unit time), position $x = ?$
$$v = \frac{dx}{dt}$$
$$dx = v \, dt$$
Substituting the given expression for $v = v_0 + gt + ft^2$ :
$$dx = (v_0 + gt + ft^2) dt$$
To find the total displacement at $t = 1\text{ s}$, integrate both sides of the equation from the initial limits ($t = 0, x = 0$) to the final limits ($t = 1, x = x$) :
$$\int_{0}^{x} dx = \int_{0}^{1} (v_0 + gt + ft^2) dt$$
$$[x]_{0}^{x} = \left[ v_0t + g\frac{t^2}{2} + f\frac{t^3}{3} \right]_{0}^{1}$$
$$x – 0 = \left( v_0(1) + \frac{g(1)^2}{2} + \frac{f(1)^3}{3} \right) – (0)$$
$$x = v_0 + \frac{g}{2} + \frac{f}{3}$$
The total displacement of the particle after unit time is $v_0 + \dfrac{g}{2} + \dfrac{f}{3}$.
Hence, the correct option is (B).
JEE Main Question
A particle located at $x = 0$ at time $t = 0$, starts moving along the positive x-direction with a velocity $v$ that varies as $v = \alpha\sqrt{x}$ (where $\alpha$ is a positive constant). The displacement of the particle varies with time as :
(A) $t^2$
(B) $t$
(C) $t^{1/2}$
(D) $t^3$
(AIEEE 2006)
Solution
Answer: (A)
Velocity-displacement relation : $$v = \alpha\sqrt{x}$$
Initial condition : At $t = 0$, position $x = 0$
$$v = \frac{dx}{dt}$$
$$\frac{dx}{dt} = \alpha\sqrt{x}$$
$$\frac{dx}{\sqrt{x}} = \alpha \, dt$$
$$x^{-1/2} \, dx = \alpha \, dt$$
Integrate both sides from the initial position ($x = 0$ at $t = 0$) to a general position $x$ at a future time $t$ :
$$\int_{0}^{x} x^{-1/2} \, dx = \int_{0}^{t} \alpha \, dt$$
$$\left[ \frac{x^{1/2}}{1/2} \right]_{0}^{x} = \alpha [t]_{0}^{t}$$
$$2\left[ \sqrt{x} \right]_{0}^{x} = \alpha (t – 0)$$
$$2\sqrt{x} = \alpha t$$
$$\sqrt{x} = \frac{\alpha t}{2}$$
$$x = \left(\frac{\alpha t}{2}\right)^2$$
$$x = \frac{\alpha^2}{4}t^2$$
Since $\alpha$ is a constant, the term $\frac{\alpha^2}{4}$ is also a constant. Therefore, the proportionality relation is :
$$x \propto t^2$$
The displacement of the particle is directly proportional to $t^2$.
Hence, the correct option is (A).
JEE Main Question
The relation between time $t$ and distance $x$ is $t = ax^2 + bx$, where $a$ and $b$ are constants. The acceleration is (where $v$ denotes velocity) :
(A) $-2abv^2$
(B) $2bv^3$
(C) $-2av^3$
(D) $2av^2$
(AIEEE 2005)
Solution
Answer: (C)
Time-displacement relation : $$t = ax^2 + bx $$
Differentiate both sides with respect to time ($t$) :
$$\frac{dt}{dt} = \frac{d}{dt}(ax^2 + bx)$$
$$1 = 2ax\frac{dx}{dt} + b\frac{dx}{dt}$$
Since velocity is defined as the rate of change of position with time ($v = \dfrac{dx}{dt}$), substitute $v$ into the expression :
$$1 = 2axv + bv$$
$$1 = v(2ax + b)$$
$$v = \frac{1}{2ax + b} $$
$$v^{-1} = 2ax + b $$
To find the acceleration ($A = \dfrac{dv}{dt}$), differentiate with respect to time ($t$) :
$$\frac{d}{dt}(v^{-1}) = \frac{d}{dt}(2ax + b)$$
$$-\dfrac{1}{v^2} \dfrac{dv}{dt}= 2a \frac{dx}{dt}$$
$$-\frac{1}{v^2}A = 2av$$
$$A = -2av \cdot v^2$$
$$A = -2av^3$$
The acceleration of the particle is $-2av^3$.
Hence, the correct option is (C).
JEE Main Question
A particle is moving with constant acceleration $a$. The following graph shows a $v^2$ versus $x$ (displacement) plot. The acceleration of the particle is _______ $\text{m/s}^2$.
![JEE Main PYQ [[2021, 31 Aug Shift-II] Solved Numerical Problem](https://physicsanandclasses.co.in/wp-content/uploads/2026/05/JEE-Main-2021-31-Aug-Shift-II-Solved-Problem.webp "JEE Main Motion in a Straight Line MCQs PYQs Previous Year Questions and Solutions 3")
(JEE Main 2021, 31 Aug Shift-II)
Solution
Answer: 1
Method 1: Using the Slope of a Straight Line
From the third equation of motion under constant acceleration, we have :
$$v^2 = u^2 + 2ax$$
Let’s rearrange this to match the standard equation of a straight line, $y = mx + c$ :
$$\underbrace{v^2}_{y} = \underbrace{(2a)}_{m}\cdot \underbrace{x}_{x} + \underbrace{u^2}_{c}$$
By comparing the two expressions :
The variable on the vertical axis ($y$-axis) is $v^2$.
The variable on the horizontal axis ($x$-axis) is $x$.
The slope ($m$) of the line represents $2a$.
We can calculate the slope ($m$) using any two coordinates on the straight line.
Initial point (on $y$-axis where $x = 0$) : $(0, 20)$
Final point $C$ : $(30, 80)$
The formula for the slope is :
$$\text{Slope } (m) = \frac{y_2 – y_1}{x_2 – x_1}$$
$$\text{Slope } (m) = \frac{80 – 20}{30 – 0} = \frac{60}{30} = 2$$
Since the slope equals $2a$ :
$$2a = \text{Slope}$$
$$2a = 2$$
$$a = 1\text{ m/s}^2$$
Method 2: Using the Kinematic Equation Directly
Alternatively, you can skip line equations entirely and use the equation of motion between two explicit coordinate points given on the plot, like point $A(10, 40)$ and point $C(30, 80)$.
At $x_1 = 10\text{ m}$, $v_1^2 = 40\text{ (m/s)}^2$
At $x_2 = 30\text{ m}$, $v_2^2 = 80\text{ (m/s)}^2$
Applying the equation of motion :
$$v_2^2 = v_1^2 + 2a(x_2 – x_1)$$
Substitute the coordinates :
$$80 = 40 + 2a(30 – 10)$$
$$80 – 40 = 2a(20)$$
$$40 = 40a$$
$$a = 1\text{ m/s}^2$$
The acceleration of the particle is $1\text{ m/s}^2$.
JEE Main Question
Water drops are falling from a nozzle of a shower onto the floor from a height of $9.8\text{ m}$. The drops fall at a regular interval of time. When the first drop strikes the floor, at that instant, the third drop begins to fall. Locate the position of the second drop from the floor when the first drop strikes the floor.
(A) $4.18\text{ m}$
(B) $2.94\text{ m}$
(C) $2.45\text{ m}$
(D) $7.35\text{ m}$
(JEE Main 2021, 27 Aug Shift-II)
Solution
Answer: (D)
Let the fixed time interval between any two consecutive drops be $\Delta t$.
When the 1st drop hits the floor, the 3rd drop is just leaving the nozzle.
This means : Time spent in the air by the 1st drop = $2\Delta t$ and Time spent in the air by the 2nd drop = $\Delta t$
The total height is $H = 9.8\text{ m}$ and the initial velocity of the drops is $u = 0$. Using the second equation of motion ($H = ut + \frac{1}{2}gt^2$) :
$$H = \frac{1}{2}g T^2$$
Taking $g = 9.8\text{ m/s}^2$ :
$$9.8 = \frac{1}{2} \times 9.8 \times T^2$$
$$1 = \frac{1}{2} T^2 \implies T^2 = 2 \implies T = \sqrt{2}\text{ s}$$
Since the total time of flight for the first drop is $T = 2\Delta t$ :
$$2\Delta t = \sqrt{2}$$
$$\Delta t = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\text{ s}$$
The second drop has been falling freely from rest for a duration of exactly $t = \Delta t$ :
$$x = \frac{1}{2}g(\Delta t)^2$$
Substitute the values of $g$ and $\Delta t$ :
$$x = \frac{1}{2} \times 9.8 \times \left(\frac{1}{\sqrt{2}}\right)^2$$
$$x = \frac{1}{2} \times 9.8 \times \frac{1}{2}$$
$$x = \frac{9.8}{4} = 2.45\text{ m}$$
So, the second drop has fallen a distance of $2.45\text{ m}$ downwards from the nozzle.
The question asks for the position of the second drop measured from the floor :
$$\text{Height from floor} = H – x$$
$$\text{Height from floor} = 9.8\text{ m} – 2.45\text{ m} = 7.35\text{ m}$$
The position of the second drop from the floor is $7.35\text{ m}$.
Hence, the correct option is (D).
JEE Main Question
Two spherical balls having equal masses with a radius of 5 cm each are thrown upwards along the same vertical direction at an interval of 3 s with the same initial velocity of 35 m/s. These balls collide at a height of _______ m. (Take g = 10m/s2)
(JEE Main 2021, 26 Aug Shift-I)
Solution
Answer: 50
Let the first ball be thrown at time t = 0.
Since the second ball is thrown after an interval of 3 s, the time experienced by each ball when they collide at a total elapsed time t will be :
Time of flight for the 1st ball = t
Time of flight for the 2nd ball = t – 3
Both balls are projected upward with an initial velocity $u$ = 35 m/s under a downward acceleration due to gravity g = 10 m/s2.
At the instant of collision, both balls are at the exact same vertical position (height h) from the point of projection. Using the second equation of motion ($h = ut – \frac{1}{2}gt^2$) :
For the 1st ball :
$$h = 35t – \frac{1}{2}(10)t^2 \implies h = 35t – 5t^2 $$
For the 2nd ball :
$$h = 35(t – 3) – \frac{1}{2}(10)(t – 3)^2 \implies h = 35(t – 3) – 5(t – 3)^2 $$
Equating both above equations, we get :
$$35t – 5t^2 = 35(t – 3) – 5(t – 3)^2$$
$$35t – 5t^2 = 35t – 105 – 5(t^2 – 6t + 9)$$
$$35t – 5t^2 = 35t – 105 – 5t^2 + 30t – 45$$
$$0 = -105 + 30t – 45$$
$$0 = 30t – 150$$
$$30t = 150$$
$$t = \frac{150}{30} = 5\text{ s}$$
This means the collision occurs exactly 5 seconds after the first ball is thrown.
Substitute the value of $t = 5 s back into above Equation :
$$h = 35(5) – 5(5)^2$$
$$h = 175 – 5(25)$$
$$h = 175 – 125$$
$$h = 50\text{ m}$$
(Note: The radius of the balls (5 cm = 0.05 m) is small enough compared to 50 m that it does not affect the macroscopic kinematic calculation).
The two balls collide at a height of 50 m.
JEE Main Question
From a building, two balls A and B are thrown such that A is thrown upwards and B downwards (both vertically) with the same initial speed. If $v_A$ and $v_B$ are their respective velocities on reaching the ground, then :
(a) $v_B > v_A$
(b) $v_A = v_B$
(c) $v_A > v_B$
(d) their velocities depend on their masses
(AIEEE 2002)
Solution
Answer: (b)
This problem can be elegantly solved using two different approaches: Kinematics and the Law of Conservation of Energy.
Method 1: Using the Law of Conservation of Energy
Let the height of the building be h, and both balls be thrown with the same initial speed $u$. Let the mass of ball A be $m_A$ and the mass of ball B be $m_B$.
Total energy of a ball at the top of the building is the sum of its potential energy and kinetic energy. Total energy at the ground is purely kinetic energy.
For Ball A : $$\text{Total Energy at top} = \text{Total Energy at ground}$$$$m_A gh + \frac{1}{2}m_A u^2 = \frac{1}{2}m_A v_A^2$$Dividing by mass $m_A$:$$gh + \frac{1}{2}u^2 = \frac{1}{2}v_A^2 \implies v_A = \sqrt{u^2 + 2gh}$$
For Ball B : $$\text{Total Energy at top} = \text{Total Energy at ground}$$$$m_B gh + \frac{1}{2}m_B u^2 = \frac{1}{2}m_B v_B^2$$Dividing by mass $m_B$:$$gh + \frac{1}{2}u^2 = \frac{1}{2}v_B^2 \implies v_B = \sqrt{u^2 + 2gh}$$
Comparing the two results, we find :
$$v_A = v_B$$
Method 2: Using Kinematics (Sign Conventions)
Let’s choose the downward direction as positive (+).
Acceleration for both balls is due to gravity: $a = +g$
Net displacement for both balls from the top to the ground is downward : s = +h
Now let’s look at their initial velocities :
Ball B is thrown directly downwards : $u_B = +u$$$v_B^2 = (+u)^2 + 2gh = u^2 + 2gh$$
Ball A is thrown upwards : $u_A = -u$$$v_A^2 = (-u)^2 + 2gh = u^2 + 2gh$$
Since $(-u)^2 = u^2$, the squares of their final velocities are identical :
$$v_A^2 = v_B^2 \implies v_A = v_B$$
(Physically, when Ball $A$ reaches its highest point and falls back down to the level of the roof, it passes the roof line going downwards with the exact same speed $u$ that Ball $B$ started with).
The final velocities of both balls when they strike the ground are completely identical and independent of their paths or masses.
Hence, the correct option is (b).