Q1JEE Main 2025MCQ

The velocity a time graph of an object moving along a straight line is shown in figure. What is the distance covered by the object between ${\text{t}}=0$ to ${\text{t}}=$ $4 s$ ?[28 Jan 2025 Shift 2]

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📖 Explanation

Distance travelled = displacement when direction of velocity remains constant $\Rightarrow \;\text{ Distance }\;\;=\;\text{ Area }\;$ $\;=\;\frac{1}{2}(2 s+4 s)(10 m / s)$ $\;=30 m$

Q2JEE Main 2025MCQ

The motion of an airplane is represented by velocity-time graph as shown below. The distance covered by airplane in the first 30.5 second is______ km. [23 Jan 2025 Shift 1]

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📖 Explanation

Distance = area under the graph $d\;=300 \times 2+400 \times 28.5$ $\;=600+114000$ $\;=12000 {\text{m}}$

Q3JEE Main 2024NAT

A particle initially at rest starts moving from reference point. ${\text{x}}=0$ along ${\text{x}}$ -axis, with velocity $v$ that varies as $v=4 \sqrt{ {\text{x}}} {\text{m}} / {\ s}$ . The acceleration of the particle is____ ${\text{ms}}^{-2}$ . [1-Feb-2024 Shift 2]

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📖 Explanation

$\; {\text{V}}=4 \sqrt{ {\text{x}}}$ $\; {\text{a}}= {\text{V}} \;{ {\text{dv}}}/{ {\text{dx}}}$ $\;=4 \sqrt{ {\text{x}}} \times 4 \times \;\frac{1}{2} {\text{x}}^{-1 / 2}$ $=8 {\text{m}} / {\ s}^2$

Q4JEE Main 2024MCQ

Two cars are travelling towards each other at speed of $20 {\text{m}} {\ s}^{-1}$ each. When the cars are $300 {\text{m}}$ apart, both the drivers apply brakes and the cars retard at the rate of $2 {\text{m}} {\ s}^{-2}$ . The distance between them when they come to rest is :[9 Apr 2024 Shift 2]

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📖 Explanation

$\;{| { {\text{u}}}^{\to }_{ {\text{BA}}} |}=40 {\text{m}} / {\ s}$ $\;{| { {\text{a}}}^{\to }_{ {\text{BA}}} |}=4 {\text{m}} / {\ s}$ $\;\;\text{ Apply }\; ( {\text{v}}^2= {\text{u}}^{2+}2 {\text{as}}) _{\;\text{relative }\;}$ $\; {\text{O}}=(40)^{2+}2(-4)( {\text{S}})$ $\; {\text{S}}=200 {\text{m}}$ $\;\;\text{ Remaining distance }\;=300-200=100 {\text{m}}$

Q5JEE Main 2024MCQ

Train A is moving along two parallel rail tracks towards north with speed $72 {\text{km}} / {\text{h}}$ and train B is moving towards south with speed $108 {\text{km}} / {\text{h}}$ . Velocity of train B with respect to A and velocity of ground with respect to ${\text{B}}$ are (in ${\text{ms}}^{-1}$ ) : [1-Feb-2024 Shift 2]

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📖 Explanation

< $\; {\text{V}}_{ {\text{A}}}=20 {\text{m}} / {\ s}$ $\; {\text{V}}_{ {\text{B}}}=-30 {\text{m}} / {\ s}$ Velocity of B w.r.t. A ${\text{V}}_{ {\text{B}} / {\text{A}}}=-50 {\text{m}} / {\ s}$ Velocity of ground w.r.t. B ${\text{V}}_{ {\text{G}} / {\text{B}}}=30 {\text{m}} / {\ s}$ br>

Q6JEE Main 2024MCQ

A body starts moving from rest with constant acceleration covers displacement $S_1$ in first ( $p-1$ ) seconds and ${\text{S}}_2$ in first ${\text{p}}$ seconds. The displacement ${\text{S}}_1+ {\text{S}}_2$ will be made in time : [29-Jan-2024 Shift 1]

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📖 Explanation

$S_1$ in first $(p-1) {\ sec}$ ${\text{S}}_2$ in first ${\text{p}}$ sec $\; {\text{S}}_1=\;\frac{1}{2} {\text{a}}( {\text{p}}-1)^2$ $\; {\text{S}}_2=\;\frac{1}{2} {\text{a}}( {\text{p}})^2$ $\; {\text{S}}_1+ {\text{S}}_2=\;\frac{1}{2} {\text{at}}^2$ $\;( {\text{p}}-1)^{2+} {\text{p}}^2= {\text{t}}^2$ $\; {\text{t}}=\sqrt{2 {\text{p}}^{2+}1-2 {\text{p}}}$

Q7JEE Main 2024NAT

A particle moves in a straight line so that its displacement $x$ at any time $t$ is given by $x^2=1+t^2$ . Its acceleration at any time $t$ is $x^{-n}$ where $n=$ ______.[6 Apr 2024 Shift 2]

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📖 Explanation

$\;x^2=1+t^2$ $\;2 x \;\frac{d x}{d t}=2 t$ $\;x v=t$ $\;x \;\frac{d v}{d t}+v \;\frac{d x}{d t}=1$ $\;x \cdot a+v^2=1$ $\;a=\;\frac{1-v^2}{x}=\;\frac{1-t^2 / x^2}{x}$ $\;a=\;\frac{1}{x^3}=x^{-3}$

Q8JEE Main 2024NAT

The displacement and the increase in the velocity of a moving particle in the time interval of $t$ to $(t+1) {\ s}$ are $125 {\text{m}}$ and $50 {\text{m}} / {\ s}$ , respectively. The distance travelled by the particle in $( {\text{t}}+2)^{ {\text{th}}} {\ s}$ is ____ ${\text{m}}$ . [30-Jan-2024 Shift 1]

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📖 Explanation

Considering acceleration is constant $\;v=u+a t$ $\;u+50=u+a \Rightarrow a=50 {\text{m}} / {\ s}^2$ $\;125=u t+\;\frac{1}{2} a t^2$ $\;125=u+\;\frac{a}{2}$ $\;\Rightarrow u=100 {\text{m}} / {\ s}$ $\; \therefore S_{n^{t h}}=u+\;\frac{a}{2}[2 n-1]$ $\;=175 {\text{m}}$

Q9JEE Main 2024MCQ

The relation between time ' $t$ ' and distance ' $x$ ' is $t=$ $\alpha x^{2+}\beta x$ , where $\alpha$ and $\beta$ are constants. The relation between acceleration (a) and velocity (v) is: [31-Jan-2024 Shift 1]

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📖 Explanation

${\text{t}}=\alpha {\text{x}}^{2+}\beta {\text{x}}$ (differentiating wrt time) $\;\;{ {\text{dt}}}/{ {\text{dx}}}=2 \alpha {\text{x}}+\beta$ $\;\;{1}/{ {\text{v}}}=2 \alpha {\text{x}}+\beta$ (differentiating wrt time) $\;-\;\frac{1}{v^2} \;\frac{d v}{d t}=2 \alpha \;\frac{d x}{d t}$ $\;\;\frac{d v}{d t}=-2 \alpha v^3$

Q10JEE Main 2024NAT

A body falling under gravity covers two points A and $B$ separated by $80 {\text{m}}$ in $2 {\ s}$ . The distance of upper point A from the starting point is____ ${\text{m}} (.$ use ${\text{g}}=10 {\text{ms}}^{-2})$ [27-Jan-2024 Shift 2]

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📖 Explanation

From $A \longrightarrow B$ $\;-80=- {\text{v}}_1 {\text{t}}-\;\frac{1}{2} \times 10 {\text{t}}^2$ $\;-80=-2 {\text{v}}_1-\;\frac{1}{2} \times 10 \times 2^2$ $\;-80=-2 {\text{v}}_1-20$ $\;-60=-2 {\text{v}}_1$ $\; {\text{v}}_1=30 {\text{m}} / {\ s}$ From ${\text{O}}$ to ${\text{A}}$ $\; {\text{v}}^2= {\text{u}}^{2+}2 {\text{gS}}$ $\;30^2=0+2 \times (-10)(- {\text{S}})$ $\;900=20 {\text{S}}$ $\; {\text{S}}=45 {\text{m}}$

Q11JEE Main 2024MCQ

A particle is moving in a straight line. The variation of position ' $x$ ' as a function of time ' $t$ ' is given as $x= (t^{3-}6 t^{2+}20 t+15) {\text{m}}$ . The velocity of the body when its acceleration becomes zero is : [29-Jan-2024 Shift 2]

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📖 Explanation

$\; {\text{x}}= {\text{t}}^{3-}6 {\text{t}}^{2+}20 {\text{t}}+15$ $\;\;{ {\text{dx}}}/{ {\text{dt}}}= {\text{v}}=3 {\text{t}}^{2-}12 {\text{t}}+20$ $\;\;{ {\text{dv}}}/{ {\text{dt}}}= {\text{a}}=6 {\text{t}}-12$ $\;\;\text{ When }\; {\text{a}}=0$ $\;6 {\text{t}}-12=0 ; {\text{t}}=2 {\ sec}$ $\;\;\text{ At }\; {\text{t}}=2 {\ sec}$ $\; {\text{v}}=3(2)^{2-}12(2)+20$ $\; {\text{v}}=8 {\text{m}} / {\ s}$

Q12JEE Main 2024NAT

A body moves on a frictionless plane starting from rest. If $S_n$ is distance moved between $t=n-1$ and $t$ $= {\text{n}}$ and ${\text{S}}_{ {\text{n}}-1}$ is distance moved between ${\text{t}}= {\text{n}}-2$ and $t=n-1$ , then the ratio $\;\frac{S_{n-1}}{S_n}$ is $(1-\;\frac{2}{x})$ for $n$ $=10$ . The value of $x$ is _______[5 Apr 2024 Shift 1]

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📖 Explanation

$\; {\text{S}}_{ {\text{n}}}=\;\frac{1}{2} {\text{a}}(2 {\text{n}}-1)=\;{19 {\text{a}}}/{2}$ $\; {\text{S}}_{ {\text{n}}-1}=\;\frac{1}{2} {\text{a}}(2 {\text{n}}-3)=\;{17 {\text{a}}}/{2}$ $\;\;{ {\text{S}}_{ {\text{n}}-1}}/{ {\text{S}}_{ {\text{n}}}}=\;\frac{17}{19}=1-\;{2}/{ {\text{x}}} \Rightarrow {\text{x}}=19$

Q13JEE Main 2024MCQ

A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in $t_1$ . If it is projected vertically downwards from the same point with the same speed, it reaches the ground in $t_2$ . Time required to reach the ground, if it is dropped from the top of the tower, is :[6 Apr 2024 Shift 2]

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📖 Explanation

$\;t_1=\;{u+{\sqrt{u^{2+}2 g h}}}/{g}$ $\;t_2=\;{-u+{\sqrt{u^{2+}2 g h}}}/{g}$ $\;t=\;{{\sqrt{2 g h}}}/{g}$ $\;t_1 t_2=\;\frac{ (u^{2+}2 g h) -u^2}{g^2}=\;\frac{2 g h}{g^2}=t^2$ $\;\Rightarrow t={\sqrt{t_1 t_2}}$

Q14JEE Main 2024MCQ

A train starting from rest first accelerates uniformly up to a speed of $80 {\text{km}} / {\text{h}}$ for time $t$ , then it moves with a constant speed for time 3t. The average speed of the train for this duration of journey will be (in ${\text{km}} / {\text{h}}$ ) :[6 Apr 2024 Shift 1]

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📖 Explanation

$\;\text{ Average speed }\;\;=\;\frac{\;\text{ total distance }\;}{\;\text{ time taken }\;}$ $\;=\;{\;{80 \times {\text{t}}}/{2}+80 \times 3 {\text{t}}}/{4 {\text{t}}}=70 {\text{km}} / {\text{hr}} .$

Q15JEE Main 2024MCQ

A particle moving in a straight line covers half the distance with speed $6 {\text{m}} / {\ s}$ . The other half is covered in two equal time intervals with speeds 9 ${\text{m}} / {\ s}$ and $15 {\text{m}} / {\ s}$ respectively. The average speed of the particle during the motion is :[9 Apr 2024 Shift 1]

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📖 Explanation

$B D \Rightarrow S\;=9 t+15 t=24 t$ $A B \Rightarrow S\;=6 t_1=24 t \Rightarrow t_1=4 t$ $< \;\text{ speed }\; > \;=\;\frac{\;\text{ dist. }\;}{\;\text{ time }\;}=\;\frac{48 t}{2 t+t_1}$ $\;=\;\frac{48 t}{2 t+4 t} \Rightarrow \;\frac{48 t}{6 t} \Rightarrow 8 {\text{m}} / {\ s}$

Q16JEE Main 2024MCQ

A body travels $102.5 {\text{m}}$ in ${\text{n}}^{\;\text{th }\;}$ second and $115.0 {\text{m}}$ in $( {\text{n}}+2)^{\;\text{th }\;}$ second. The acceleration is :[4 Apr 2024 Shift 1]

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📖 Explanation

Given, $102.5= {\text{u}}+\;{ {\text{a}}}/{2}(2 {\text{n}}-1) &$ $\;115= {\text{u}}+\;{ {\text{a}}}/{2}(2 {\text{n}}+3)$ $\;\Rightarrow 102.5= {\text{u}}+ {\text{an}}-\;{ {\text{a}}}/{2} &$ $\;115= {\text{u}}+ {\text{an}}+\;{3 {\text{a}}}/{2}$ $\;12.5=2 {\text{a}} \Rightarrow {\text{a}}=6.25 {\text{m}} / {\ s}^2$ Hence option (2)

Q17JEE Main 2024NAT

A bus moving along a straight highway with speed of $72 {\text{km}} / {\text{h}}$ is brought to halt within $4 {\ s}$ after applying the brakes. The distance travelled by the bus during this time (Assume the retardation is uniform) is ________ ${\text{m}}$ .[4 Apr 2024 Shift 2]

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📖 Explanation

Initial velocity $= {\text{u}}=72 {\text{km}} / {\text{h}}=20 {\text{m}} / {\ s}$ $\; {\text{v}}= {\text{u}}+ {\text{at}}$ $\;\Rightarrow 0=20+ {\text{a}} \times 4$ $\; {\text{a}}=-5 {\text{m}} / {\ s}^2$ $\; {\text{v}}^{2-} {\text{u}}^2=2 {\text{as}}$ $\;\Rightarrow 0^{2-}20^2=2(-5) . {\ s}$ $\; {\ s}=40 {\text{m}}$

Q18JEE Main 2024NAT

A particle is moving in one dimension (along ${\text{x}}$ axis) under the action of a variable force. It's initial position was $16 {\text{m}}$ right of origin. The variation of its position (x) with time (t) is given as $x=-3 t^{3+}18 t^{2+}16 t$ , where $x$ is in $m$ and $t$ is in $s$ . The velocity of the particle when its acceleration becomes zero is_______ ${\text{m}} / {\ s}$ . [1-Feb-2024 Shift 1]

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📖 Explanation

$\;x=3 t^{3+}18 t^{2+}16 t$ $\;v=-9 t^{2+}36+16$ $\;a=-18 t+36$ $\;a=0 \;\text{ at t }\;=2 s$ $\;v=-9(2)^{2+}36 \times 2+16$ $\;v=52 {\text{m}} / {\ s}$

Q19JEE Main 2023MCQ

The velocity time graph of a body moving in a straight line is shown in figure.The ratio of displacement to distance travelled by the body in time 0 to $10 {\ s}$ is [24-Jan-2023 Shift 2]

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📖 Explanation

Displacement $=$ area $=16-8+16-8=16 {\text{m}}$ Distance $=Σ \mid$ area $\mid =48 {\text{m}}$ $\;\frac{\;\text{ displacement }\;}{\;\text{ Distance }\;}=\;\frac{1}{3}$

Q20JEE Main 2023MCQ

A ball is thrown vertically upward with an initial velocity of $150 {\text{m}} / {\ s}$ . The ratio of velocity after $3 {\ s}$ and $5 {\ s}$ is $\;\frac{x+1}{x}$ . The value of $x$ is _______. {take, $g=10 {\text{m}} / {\ s}^2$ } [12-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ ToF }\;=\;\frac{24}{g}=\;\frac{2 \times 150}{10}=30 {\ sec}$ $\;\;\text{ ToA }\;=15 {\ sec} .$ $\;\;{ {\text{V}}_1}/{ {\text{V}}_2}=\;{ {\text{u}}+ {\text{at}}_1}/{ {\text{u}}+ {\text{at}}_2}$ $\;\;{ {\text{V}}_1}/{ {\text{V}}_2}=\;\frac{150-10 \times 3}{150-10 \times 5}=\;\frac{120}{100}$ $\;\;{ {\text{V}}_1}/{ {\text{V}}_2}=\;\frac{6}{5}$ $\; {\text{x}}+1=6$ $\; {\text{x}}=5$

Motion In A Straight Line Practice Questions

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