JEE Main 2019 Kinematics Lecture Flashcards

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A set of vocabulary flashcards covering key concepts, formulas, and problems from the JEE Main 2019 Kinematics lecture.

Last updated 1:51 AM on 9/14/26
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22 Terms

1
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Position from Velocity-Time Graph

The position or displacement of a particle starting from the origin calculated by evaluating the area under the velocity-time graph up to time tt.

2
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Relative Velocity of Trains

The magnitude of relative speed between two trains of speeds v1v_1 and v2v_2, defined as v1−v2v_1 - v_2 when traveling in the same direction and v1+v2v_1 + v_2 when traveling in opposite directions.

3
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Uniform Acceleration Motion Graphs

Graphical curves for constant acceleration motion starting from rest: acceleration vs. time is a horizontal line, velocity vs. time is a straight line through the origin, and position vs. time is a parabola opening upwards from the origin.

4
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Momentum-Height Relation in Vertical Throw

The quadratic relation p2=m2u02−2gm2hp^2 = m^2 u_0^2 - 2 g m^2 h describing momentum pp as a function of height hh for a projectile thrown vertically upwards with initial speed u0u_0.

5
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Zero Acceleration Point

The time instant t=b3ct = \frac{b}{3 c} at which acceleration equals zero for position function x(t)=at+bt2−ct3x(t) = a t + b t^2 - c t^3, giving velocity v=a+b23cv = a + \frac{b^2}{3 c}.

6
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Relative Overtaking Speed Formula in Car Race

The relation v = \root\frac{}{a_1 a_2} t giving the additional speed vv with which Car A crosses the finish line relative to Car B when starting from rest with accelerations a1a_1 and a2a_2 and finishing time difference tt.

7
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Speed in 3D Motion

The total speed of a particle moving in 3D space given by v = \root\frac{}{v_x^2 + v_y^2 + v_z^2}, derived by differentiating individual spatial coordinates with respect to time.

8
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Acceleration from Velocity-Position Function

The uniform acceleration a=vdvdx=b22a = v \frac{dv}{dx} = \frac{b^2}{2} obtained from a velocity function v = b \root\frac{}{x}.

9
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Trajectory Equation for Coupled Velocities

The parabolic or hyperbolic path y2=x2+Cy^2 = x^2 + C obtained by integrating the differential relation dydx=vyvx=xy\frac{dy}{dx} = \frac{v_y}{v_x} = \frac{x}{y} for velocity components vx=kyv_x = k y and vy=kxv_y = k x.

10
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Maximum Coverage Area of Projectile Guns

The maximum ground area Area=θ×range2const=θu4g2\text{Area} = \frac{\theta \times \text{range}^2}{\text{const}} = \frac{\theta u^4}{g^2} covered on a 2D plane by firing projectiles in all horizontal directions at a 45o45^\text{o} angle, which scales as u4u^4.

11
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Time of Flight on an Inclined Plane

The duration T=2utan⁡(θ)gtan⁡(θ)tan⁡(θ)T = \frac{2 u \tan(\theta)}{g \tan(\theta) \tan(\theta)} or T=2utan⁡(θ)gtan⁡(θ)T = \frac{2 u \tan(\theta)}{g \tan(\theta)} given by T=2uyay=2utan⁡(θ)gtan⁡(θ)T = \frac{2 u_y}{a_y} = \frac{2 u \tan(\theta)}{g \tan(\theta)}, specifically written as T=2utan⁡(θ)gtan⁡(θ)T = \frac{2 u \tan(\theta)}{g \tan(\theta)} along an incline of angle tan⁡(θ)\tan(\theta).

12
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Range on an Inclined Plane

The total distance along an incline of slope angle tan⁡(θ)\tan(\theta) hit by a projectile launched at relative angle tan⁡(θ)\tan(\theta), given by R=2u2tan⁡(θ)tan⁡(θ+tan⁡(θ))gtan⁡2(tan⁡(θ))R = \frac{2 u^2 \tan(\theta) \tan(\theta + \tan(\theta))}{g \tan^2(\tan(\theta))}.

13
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Zenith Time with Quadratic Drag Force

The time t = \frac{1}{\root\frac{}{\tan(\theta) g}} \tan^{-1}\tan(v_0 \root\frac{}{\frac{\tan(\theta)}{g}}\tan) taken by a particle launched upward with speed v0v_0 to reach zero speed under gravity and air resistance mtan⁡(θ)v2m \tan(\theta) v^2.

14
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Product of Times of Flight for Same Range

The product T1T2=2RgT_1 T_2 = \frac{2 R}{g} for two complementary launch angles tan⁡(θ)\tan(\theta) and 90o−tan⁡(θ)90^\text{o} - \tan(\theta) yielding equal horizontal range RR.

15
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Standard Equation of Projectile Trajectory

The mathematical formula y=xtan⁡(tan⁡(θ)0)−gx22u2tan⁡2(tan⁡(θ)0)y = x \tan(\tan(\theta)_0) - \frac{g x^2}{2 u^2 \tan^2(\tan(\theta)_0)} describing the parabolic path of a projectile.

16
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Product of Maximum Heights

The relationship R2=16H1H2R^2 = 16 H_1 H_2 connecting horizontal range RR to maximum heights H1H_1 and H2H_2 reached at complementary launch angles.

17
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Radius of Curvature

The radius R=v2a⊥R = \frac{v^2}{a_\bot} of the osculating circle along a curved motion trajectory, where a⊥a_\bot is the perpendicular component of acceleration relative to velocity vv.

18
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Change in Velocity Vector for Circular Motion

The magnitude of change in velocity tan⁡(θ)v=2vtan⁡(tan⁡(θ)2)\tan(\theta) v = 2 v \tan(\frac{\tan(\theta)}{2}) when a particle moves at constant speed vv through a central rotation angle tan⁡(θ)\tan(\theta).

19
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Relative Velocity in Concentric Circles

The relative velocity vector tan⁡(v)A−tan⁡(v)B\tan(v)_A - \tan(v)_B between two particles orbiting concentric circular tracks of radii R1R_1 and R2R_2 at equal angular velocity tan⁡(θ)\tan(\theta).

20
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Apparent Sound Angle and Plane Speed

The condition vplane=vsoundtan⁡(tan⁡(θ))v_{\text{plane}} = v_{\text{sound}} \tan(\tan(\theta)) where a horizontally traveling jet plane passes directly overhead as sound emitted at angle tan⁡(θ)\tan(\theta) reaches an observer on the ground.

21
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Minimum Separation Time in Relative Motion

The time t=tan⁡(r)×tan⁡(v)∣tan⁡(v)∣2t = \frac{\tan(r) \times \tan(v)}{|\tan(v)|^2} at which two moving ships achieve minimum distance of approach, calculated using initial relative position vector tan⁡(r)\tan(r) and relative velocity vector tan⁡(v)\tan(v).

22
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Upstream Angle for Direct River Crossing

The heading angle tan⁡(θ)=120o\tan(\theta) = 120^\text{o} relative to river flow required for a swimmer of speed 4 km/h4\text{ km/h} in still water to cross directly across a river flowing at 2 km/h2\text{ km/h}.