Kinematics 1D- JEE Main Previous Year Questions with Solutions

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Simulator Previous Years AIEEE/JEE Mains Questions [esquestion] A particle has an initial velocity of $3 \hat{i}+4 \hat{j}$ and an acceleration of $0.4 \hat{\mathrm{i}}+0.3 \hat{\mathrm{j}}$. Its speed after 10s is:- (1) 7 units (2) 8.5 units (3) 10 units (4) $7 \sqrt{2}$ units #tag# [AIEEE-2009] #sol# (4) v = u + at u = 3i + 4j + (0.4 i + 0.3 j) × 10 = 3i + 4j + 4i + 3i u = 7i + 7j $|\overrightarrow{\mathrm{u}}|=\sqrt{7^{2}+7^{2}}$ $=7 \sqrt{2}$ [/esquestion] [esquestion] An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by $\frac{\mathrm{dv}}{\mathrm{dt}}=-2.5 \sqrt{\mathrm{v}}$ where v is the instantaneous speed. The time taken by the object, to come to rest, would be :- (1) 4 s (2) 8 s (3) 1 s (4) 2 s #tag# [AIEEE-2011] #sol# (4) $\overrightarrow{\mathrm{a}}=\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}$ $\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}=-2.5 \sqrt{\mathrm{v}}$ $\int \mathrm{d} \mathrm{v} \times \mathrm{v}^{-1 / 2}=\int-2.5 \mathrm{dt}$ $\frac{\mathrm{v}^{\frac{1}{2}}}{\frac{1}{2}}=-2.5 \mathrm{t}$ $2 \sqrt{\mathrm{v}}=-2.5 \mathrm{t}+\mathrm{c}$ $2 \sqrt{6.25}=-2.5 \mathrm{t}+\mathrm{c}$ 2 × 0.25 = c 5 = c $2 \sqrt{\mathrm{v}}=-\mathrm{kt}+5$ 0 = –kt + 5 kt = 5 $\mathrm{t}=\frac{5 \mathrm{sec}}{2.5}=2 \mathrm{sec}$ [/esquestion] [esquestion] From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is : (1) $2 \mathrm{g} \mathrm{H}=\mathrm{nu}^{2}(\mathrm{n}-2)$ (2) $\mathrm{g} \mathrm{H}=(\mathrm{n}-2) \mathrm{u}^{2}$ (3) $2 \mathrm{g} \mathrm{H}=\mathrm{n}^{2} \mathrm{u}^{2}$ (4) $\mathrm{g} \mathrm{H}=(\mathrm{n}-2)^{2} \mathrm{u}^{2}$ #tag# [JEE-MAIN-2014] #sol# (1) Time to reach highest point $=\mathrm{t}=\frac{\mathrm{u}}{\mathrm{g}}$ time to reach ground = nt $\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$ $-\mathrm{H}=\mathrm{u}(\mathrm{nt})-\frac{1}{2} \mathrm{g}(\mathrm{nt})^{2}$ $\Rightarrow 2 \mathrm{gH}=\mathrm{nu}^{2}(\mathrm{n}-2)$ [/esquestion] [esquestion] A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time ? #tag# [JEE Mains - 2017] #sol# (1) Velocity at any time t is given by v = u + at v = v0 + (–g)t $\mathrm{v}=\mathrm{v}_{0}-\mathrm{gt}$ $\Rightarrow$ straight line with negative slope [/esquestion] [esquestion] All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up. #tag# [JEE Mains - 2018] #sol# (1) In this question option (2) and (4) are the corresponding position - time graph and velocity –position graph of option (3) and its distance – time graph is given as Hence incorrect graph is option (1) [/esquestion]

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These Kinematics (1D) questions are pulled from real exams — see them in their original paper format on our previous years JEE Main question papers page.