JEE Physics Motion In A Straight Line Notes Set 06

Rest, Motion and Straight-Line Motion

Rest and motion are relative: a body moves with respect to an observer when its position changes relative to that observer. Motion along one straight line is called rectilinear motion, and its vectors can be handled with plus and minus signs.

Position, Displacement and Distance

Position tells where the particle is. Displacement is the change of position, a vector that can be positive, negative or zero. Distance is the length of the path actually travelled, a scalar that is never less than the magnitude of the displacement.

Velocity and Speed

Average velocity = displacement / time interval, with SI unit m/s and dimensions [LT⁻¹].
Instantaneous velocity \( v = \dfrac{dx}{dt} \) is the limit of the average velocity as the interval shrinks to zero; it is the slope of the tangent in an x-t graph.
Average speed = total path length / total time. It is never negative and is at least as large as the magnitude of the average velocity; the two are equal for motion in one direction.
For equal distances at speeds v₁ and v₂ the average speed is \( \dfrac{2v_1v_2}{v_1+v_2} \); for equal times it is \( \dfrac{v_1+v_2}{2} \).

Acceleration

Average acceleration = change in velocity / time. Instantaneous acceleration \( a = \dfrac{dv}{dt} = \dfrac{d^2x}{dt^2} = v\dfrac{dv}{dx} \). When velocity and acceleration have the same sign the speed increases; with opposite signs the particle slows down (retardation).

Equations of Uniformly Accelerated Motion

\[ v = u + at,\quad s = ut + \tfrac{1}{2}at^2,\quad v^2 = u^2 + 2as \]
\[ s = \left(\dfrac{u+v}{2}\right)t,\qquad s_n = u + \dfrac{a}{2}(2n-1) \]
Here u is the initial velocity, v the final velocity, s the displacement and \( s_n \) the displacement in the nth second. Signs matter: choose a positive direction and keep to it.

Motion Under Gravity

Near the earth the acceleration is g = 9.8 m/s² (often taken as 10 m/s²) downward. With upward positive, a = −g. A body released from a rising balloon starts with the balloon's velocity but moves with acceleration g alone.
Time of flight for a body dropped from height h is \( \sqrt{2h/g} \), and its speed on landing is \( \sqrt{2gh} \). From rest, distances in successive equal intervals are in the ratio 1 : 3 : 5 : ...

Worked Example

Example. A car brakes from 25 m/s to 15 m/s over 200 m. Find the time and the acceleration.
Average velocity = (25 + 15)/2 = 20 m/s, so t = 200/20 = 10 s.
a = (15 − 25)/10 = −1 m/s². To stop it needs a further 15 s and 112.5 m.

Graphs of Motion

1. Slope of an x-t graph is velocity; slope of a v-t graph is acceleration.
2. Area under a v-t graph is displacement (areas below the axis count negative); counting all areas as positive gives distance.
3. Area under an a-t graph is the change in velocity.
4. For constant acceleration the x-t graph is a parabola, the v-t graph a straight line and the a-t graph a horizontal line.
5. A horizontal line on an x-t graph means rest; a horizontal line on a v-t graph means constant velocity.

Non-uniform Acceleration

When acceleration varies, integrate: \( x = \int v\,dt \) and \( \Delta v = \int a\,dt \). If a depends on time, use \( dv = a(t)\,dt \); if a depends on position, use \( v\,dv = a(x)\,dx \); if a depends on velocity, use \( dv/a(v) = dt \) and then integrate again for position.

Meeting and Chasing Problems

Write the position of each body as a function of time using one origin and one positive direction, then equate positions. A chaser with speed v starting a distance d behind a body that begins from rest with acceleration a catches it only if \( v^2 \ge 2ad \).