JEE Physics Motion In A Straight Line Notes Set 08

Motion, Rest and the Observer

Mechanics studies how bodies move and why. Kinematics only describes the motion (position, velocity, acceleration) while dynamics brings in the forces. Motion is always relative: a body is moving with respect to an observer if its position changes as seen by that observer, and nothing is at absolute rest. Here the motion is rectilinear, that is, along a single straight line, so one axis and a sign convention are enough: choose a positive direction, and every vector (position, displacement, velocity, acceleration) is just a number with a sign.

Position, Displacement and Distance

Position tells where the particle is on the axis at an instant. Displacement is the change in position, \( \Delta x=x_f-x_i \); it can be positive, negative or zero and depends only on the end points. Distance is the actual length of path covered; it is a scalar, never negative, and is at least as large as the magnitude of the displacement. They are equal only when the particle never reverses its direction. Example: going once around a circle of radius \( R \) gives distance \( 2\pi R \) but displacement zero; going half way round gives distance \( \pi R \) and displacement \( 2R \).

Velocity, Speed and Acceleration

Average velocity is displacement divided by the time taken, \( v_{av}=\frac{x_f-x_i}{t_f-t_i} \). Average speed is total distance divided by total time and is never negative; it is at least the magnitude of the average velocity, with equality for motion in one direction. Instantaneous velocity is the limit of the average velocity as the interval shrinks, \( v=\frac{dx}{dt} \), and its magnitude is the instantaneous speed.
Average acceleration is the change in velocity divided by time, \( a_{av}=\frac{v_f-v_i}{t_f-t_i} \), and instantaneous acceleration is \( a=\frac{dv}{dt}=\frac{d^2x}{dt^2} \). Units: velocity m/s, acceleration m/s squared. If velocity and acceleration have the same sign the speed increases; if opposite, the body slows down (retardation).

Two averages worth remembering. If a body covers equal distances at speeds \( v_1 \) and \( v_2 \), its average speed is the harmonic mean \( \frac{2v_1v_2}{v_1+v_2} \). If it travels for equal times at \( v_1 \) and \( v_2 \), the average speed is the arithmetic mean \( \frac{v_1+v_2}{2} \). For three equal parts of the distance, \( \frac{3v_1v_2v_3}{v_1v_2+v_2v_3+v_3v_1} \).

Uniform and Uniformly Accelerated Motion

With constant velocity \( u \): \( x=x_i+ut \), an x-t line of slope \( u \), a horizontal v-t line and zero acceleration.
With constant acceleration \( a \) (initial velocity \( u \), final velocity \( v \), displacement \( s \)):
1. \( v=u+at \)
2. \( s=ut+\tfrac12at^2 \)
3. \( v^2=u^2+2as \)
4. \( s=\frac{u+v}{2}\,t \) (average velocity is the mean of the end velocities)
5. \( s_n=u+\frac a2(2n-1) \) for the displacement in the nth second.
Remember that these hold only while the acceleration is constant, and that the quantities carry signs. From rest, the distances covered in successive equal intervals are in the ratio \( 1:3:5:7\ldots \).

Motion under Gravity

Near the Earth, a freely moving body (no air resistance) has acceleration \( g \) downward whether it is rising or falling. Taking upward as positive, use \( a=-g \): the velocity falls steadily, becomes zero at the top, and then becomes negative. The time to the top is \( u/g \), the greatest height is \( u^2/2g \) and the total time to return to the starting level is \( 2u/g \), with the same speed on return. Two useful tricks: a body dropped from a rising balloon starts with the balloon's upward velocity, and a body dropped from rest falls distances in the ratio \( 1:3:5 \) in successive seconds. For a dropped body, the fall in the last second fixes the total time through \( s_n=\frac g2(2n-1) \).

Reading Graphs

In an x-t graph the slope of the tangent is the velocity; a chord's slope is the average velocity. A turning point has zero slope. In a v-t graph the slope is the acceleration and the area between the curve and the time axis is the displacement (areas below the axis count negative); take every area positive to get the distance travelled. In an a-t graph the area gives the change in velocity. For constant acceleration, x-t is a parabola, v-t a straight line and a-t a horizontal line. A parabola \( y=ax^2+bx+c \) opens upward for \( a>0 \) with minimum value \( -\frac{D}{4a} \) where \( D=b^2-4ac \).
Handy graph facts: the v-x graph of a particle starting from rest with constant acceleration is a parabola since \( v^2=2as \); a straight v-x line through the origin means \( a=v\,dv/dx\propto x \); a straight v-t line through the origin means constant acceleration.

Chasing and Meeting Problems

Write the position of each body as a function of the same time, taking care with the origin and sign, and equate. A quadratic in time may give two meeting times (the faster pursuer passes, falls behind and is passed again) or none if the discriminant is negative. For a pursuer at constant speed \( v \) a distance \( d \) behind a body starting from rest with acceleration \( a \), the catch happens only if \( v\ge\sqrt{2ad} \). For two bodies thrown or dropped along the same vertical line, subtracting the position equations removes the \( t^2 \) terms and leaves a linear equation for the meeting time.

Non-uniform Acceleration and Calculus

Displacement is \( \int v\,dt \) and change in velocity is \( \int a\,dt \). Choose the form according to what the acceleration depends on. If \( a=a(t) \), integrate \( dv=a\,dt \) and then \( dx=v\,dt \). If \( a=a(v) \), write \( \frac{dv}{a(v)}=dt \) to find velocity against time, or use \( v\,dv=a\,dx \) to find velocity against position. If \( a=a(x) \), use \( \int v\,dv=\int a(x)\,dx \). The relation \( a=v\frac{dv}{dx} \) is the key to problems where acceleration is given in terms of position or speed.

Worked Examples

Example 1. A car covers equal distances at 40 km/h and 60 km/h. Average speed \( =\frac{2(40)(60)}{100}=48 \) km/h, not 50.

Example 2. A car brakes from 30 m/s to 10 m/s over 80 m. \( a=\frac{10^2-30^2}{2(80)}=-5\ \text{m/s}^2 \), \( t=\frac{10-30}{-5}=4 \) s, and it needs a further \( \frac{10^2}{2(5)}=10 \) m to stop.

Example 3. A stone is dropped from 80 m (\( g=10 \)). Fall time \( \sqrt{2(80)/10}=4 \) s. Distance in the last second \( 80-\tfrac12(10)(3)^2=35 \) m, which also equals \( 5(2\times4-1)=35 \) m.

Example 4. A particle has \( a=6t \), starts at the origin with velocity 2 m/s. \( v=2+3t^2 \) and \( x=2t+t^3 \). At \( t=2 \) s, \( v=14 \) m/s and \( x=12 \) m.

Example 5. A particle moves with \( a=-2v \) starting at 8 m/s. Since \( v\frac{dv}{dx}=-2v \), \( \frac{dv}{dx}=-2 \) and \( v=8-2x \). It stops after \( x=4 \) m (taking infinite time, since \( v=8e^{-2t} \)).

Example 6. A car at 20 m/s is 100 m behind a scooter that starts from rest with \( 1\ \text{m/s}^2 \). Equal positions: \( \tfrac12t^2=20t-100 \), so \( t^2-40t+200=0 \) and \( t=20-10\sqrt2\approx5.9 \) s (first meeting).