JEE Physics Kinematics Notes Set 01

Motion in One Dimension

1. Distance is the length of the path (scalar). Displacement is the change in position (vector). Average speed = total distance / total time. Average velocity = displacement / time.
2. Instantaneous velocity \( v = \frac{dx}{dt} \) and instantaneous acceleration \( a = \frac{dv}{dt} = v\frac{dv}{dx} = \frac{d^2x}{dt^2} \). The rate of change of speed is \( \frac{d|\vec v|}{dt} = \frac{\vec v\cdot\vec a}{|\vec v|} \).
3. A body speeds up when v and a have the same sign and slows down when they have opposite signs. A body can have zero velocity and non-zero acceleration (the top of a vertical throw).
4. If a body covers equal distances at speeds \( v_1 \) and \( v_2 \), the average speed is \( \frac{2v_1v_2}{v_1+v_2} \). If it travels for equal times at these speeds, the average speed is \( \frac{v_1+v_2}{2} \).

Equations of Uniformly Accelerated Motion

1. \( v = u + at \), \( s = ut + \tfrac12at^2 \), \( v^2 = u^2 + 2as \), \( s = \frac{u+v}{2}t \).
2. Distance in the n-th second: \( s_n = u + \frac a2(2n-1) \). From rest the distances in successive equal intervals are in the ratio 1 : 3 : 5 : 7.
3. From rest, \( s \propto t^2 \) and \( t \propto \sqrt s \). A body falling from rest takes \( (\sqrt2 - 1) \) times the time of the first half for the second half.
4. If the motion reverses, distance and displacement differ. Split the motion at the instant v = 0 and add the magnitudes.
5. Variable acceleration: integrate. \( v = u + \int a\,dt \) and \( x = x_0 + \int v\,dt \). When a depends on position use \( v\,dv = a\,dx \). Retardation \( kv \) gives \( v = v_0e^{-kt} \) and a finite stopping distance \( v_0/k \).

Motion Under Gravity

1. Take g = 9.8 or 10 m/s² downward. For a body thrown up with speed u: time to the top \( \frac ug \), maximum height \( \frac{u^2}{2g} \), time of flight \( \frac{2u}{g} \), and the return speed equals u.
2. At the highest point v = 0 but a = g. At equal heights the speeds on the way up and down are equal. A body passes a given height at two times t₁ and t₂ with \( t_1 + t_2 = \frac{2u}{g} \).
3. Distances covered in successive seconds of a free fall from rest are in the ratio 1 : 3 : 5.
4. With air resistance as a constant retardation r: up-acceleration is \( g + r \), down-acceleration is \( g - r \), and the descent takes longer than the ascent. The return speed is smaller than the throw speed.
5. For two bodies in free fall the relative acceleration is zero, so their separation changes at a constant rate.

Graphs of Motion

1. Slope of the x-t graph is velocity. Slope of the v-t graph is acceleration. Area under the v-t graph is displacement (areas below the axis are negative). Area under the a-t graph is the change in velocity.
2. Area under the a-x graph is \( \frac12(v^2 - u^2) \).
3. A parabolic x-t graph means constant acceleration and a straight-line v-t graph. The speed-time graph is the v-t graph with the negative part reflected above the axis.
4. The distance-time graph never decreases. The slope of the speed-time graph can differ in sign from the acceleration.

Relative Motion in One and Two Dimensions

1. \( \vec v_{AB} = \vec v_A - \vec v_B \). For motion on a line, speeds subtract when the bodies move the same way and add when they move in opposite directions.
2. For two bodies moving towards each other, time to meet = initial gap / (v₁ + v₂). Two trains of lengths l₁ and l₂ take \( \frac{l_1+l_2}{\text{relative speed}} \) to cross.
3. \( |\vec v_1 - \vec v_2| \) is greatest when the velocities are antiparallel and least when they are parallel.
4. For two projectiles launched from the same point at the same time, the relative acceleration is zero, so each moves in a straight line at constant velocity as seen from the other. Equal horizontal components keep them in a vertical line, and equal vertical components keep them level.
5. A body dropped from an accelerating vehicle is seen by the ground observer to follow a parabola. In the vehicle frame it moves along a straight line.
6. Rain problems: the apparent velocity of rain is \( \vec v_{rain} - \vec v_{man} \). The man holds the umbrella along that direction. To see the rain vertical, he must move with the horizontal component of the rain velocity.

River-Boat Problems

Let the speed of the boat in still water be v, the river speed u and the width d.
1. Minimum crossing time \( t = \frac dv \): head perpendicular to the bank. The drift is \( \frac{ud}{v} \) downstream.
2. Shortest path (zero drift), possible only when v > u: head upstream at angle θ from the perpendicular with \( \sin\theta = \frac uv \). Time \( = \frac{d}{\sqrt{v^2-u^2}} \).
3. If v < u the drift is least when the boat heads at angle θ with the upstream direction where \( \cos\theta = \frac vu \).
4. The crossing time depends only on the component of the boat's velocity perpendicular to the bank. The river flow does not change it.
5. Downstream speed is v + u and upstream speed is v − u. A round trip over a distance L takes \( \frac{2vL}{v^2-u^2} \), longer than \( \frac{2L}{v} \) in still water.
6. Aircraft in wind follow the same vector triangle: ground velocity = air velocity + wind velocity.

Projectile Motion

For projection speed u at angle θ above the horizontal from level ground:
1. Components: \( u_x = u\cos\theta \) (constant) and \( u_y = u\sin\theta \) (changes at −g).
2. Time of flight \( T = \frac{2u\sin\theta}{g} \), maximum height \( H = \frac{u^2\sin^2\theta}{2g} \), range \( R = \frac{u^2\sin2\theta}{g} \).
3. The range is maximum at 45°, \( R_{max} = \frac{u^2}{g} \), and the range is the same for θ and 90° − θ. Also \( R = 4H\cot\theta \), \( \tan\theta = \frac{4H}{R} \) and \( R = 4\sqrt{H_1H_2} \) for the two complementary angles.
4. Trajectory: \( y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta} = x\tan\theta\left(1 - \frac xR\right) \). It is a parabola.
5. At the highest point the velocity is horizontal, \( u\cos\theta \), and the kinetic energy is \( K\cos^2\theta \). The velocity and acceleration are perpendicular only at that instant.
6. Speed at height y: \( v^2 = u^2 - 2gy \). When the direction of motion makes angle α with the horizontal, \( v = \frac{u\cos\theta}{\cos\alpha} \).
7. Horizontal projection from a height h: time \( \sqrt{\frac{2h}{g}} \), range \( u\sqrt{\frac{2h}{g}} \). A dropped body and a horizontally thrown body land at the same time.
8. Projection from a tower at height h with speed u at angle θ above the horizontal: solve \( -h = u\sin\theta\,t - \tfrac12gt^2 \) for the landing time.
9. On the surface of another planet R, H and T scale as 1/g.

Projectile on an Inclined Plane

1. Choose axes along and perpendicular to the plane. For a plane of inclination α, the acceleration components are \( g\sin\alpha \) along the plane (down) and \( g\cos\alpha \) perpendicular to it.
2. A body projected with speed u at angle θ above the plane (up the plane): \( T = \frac{2u\sin\theta}{g\cos\alpha} \) and range along the plane \( R = \frac{2u^2\sin\theta\cos(\theta+\alpha)}{g\cos^2\alpha} \).
3. Projection at right angles to the plane gives \( T = \frac{2u}{g\cos\alpha} \) and range along the plane \( \frac{2u^2}{g}\tan\alpha\sec\alpha \) (down the slope).
4. The maximum range up the plane is \( \frac{u^2}{g(1+\sin\alpha)} \) at \( \theta = 45^\circ - \frac\alpha2 \) with the plane.
5. If the projectile hits the plane at right angles, its velocity along the plane is zero at that instant.

Worked Examples

1. A car covers half the distance at 40 km/h and the rest at v so that the average speed is 48 km/h. Then \( \frac{2}{48} = \frac1{40} + \frac1v \), so v = 60 km/h.
2. A ball thrown up passes a window twice, 4 s apart, and returns to the ground after 8 s. The top of the flight is at 4 s so u = 40 m/s. The crossings are at 2 s and 6 s, so the height is \( 40(2) - 5(2)^2 = 60 \) m.
3. A boat with v = 5 km/h crosses a 1 km river along the shortest path in 15 min. The effective speed is 4 km/h, so \( u = \sqrt{5^2-4^2} = 3 \) km/h.
4. A projectile has the equation y = 16x − 5x²/4. The range is 16 / (5/4) = 12.8 m, and tan θ = 16.
5. A particle has a = 3t² + 2t + 2 and u = 2 m/s. Then \( v = 2 + t^3 + t^2 + 2t \), and v(2) = 18 m/s.