JEE Physics Motion In A Straight Line Notes Set 04

Distance, Displacement and Averages

1. Distance is the total path length (a scalar, never negative). Displacement is the straight-line vector from the initial to the final position and can be positive, negative or zero.
2. Distance is always greater than or equal to the magnitude of displacement; they are equal only for straight-line motion without reversal.
3. Average speed = total distance / total time. Average velocity = displacement / time, and it is zero for a round trip.
4. Equal distances at speeds \(v_1\) and \(v_2\): \(v_{av}=\dfrac{2v_1v_2}{v_1+v_2}\) (harmonic mean). Equal times at \(v_1\) and \(v_2\): \(v_{av}=\dfrac{v_1+v_2}{2}\) (arithmetic mean).
5. If a distance is split into parts \(d_1, d_2, d_3\) with speeds \(v_1, v_2, v_3\), then \(v_{av}=\dfrac{d_1+d_2+d_3}{d_1/v_1+d_2/v_2+d_3/v_3}\).

Instantaneous Quantities and Calculus

1. \(v=\dfrac{dx}{dt}\), \(a=\dfrac{dv}{dt}=\dfrac{d^2x}{dt^2}=v\dfrac{dv}{dx}\).
2. Displacement is the area under the v-t graph, and change in velocity is the area under the a-t graph. Slope of the x-t graph gives velocity and slope of the v-t graph gives acceleration.
3. Velocity is maximum or minimum where the acceleration is zero; the particle momentarily stops where \(v=0\).
4. Average acceleration \(=\dfrac{\Delta\vec v}{\Delta t}\). Velocity is a vector, so a change in direction at constant speed is an acceleration.
5. When a is given as a function of x, use \(v\,dv=a\,dx\); when a is a function of t, integrate directly with respect to t.

Equations of Uniformly Accelerated Motion

From rest, the distances covered in successive equal intervals are in the ratio 1 : 3 : 5 : 7 : ... and the distance covered in the first n seconds is proportional to \(n^2\). For braking with the same retardation, the stopping distance is proportional to the square of the initial speed.

Free Fall and Vertical Throw

1. Take \(a=g\) downwards. Time of fall from height h is \(t=\sqrt{2h/g}\), and the landing speed is \(\sqrt{2gh}\); both are independent of mass.
2. For a vertical throw with speed u: maximum height \(H=\dfrac{u^2}{2g}\), time of ascent \(=\dfrac ug\), total time of flight \(=\dfrac{2u}g\).
3. The motion is symmetric: the speed at a given height is the same going up and coming down, and the time to rise from a point to the top equals the time to fall back to it.
4. The distance covered in the last t seconds of the rise is \(\tfrac12gt^2\) (reverse the motion and treat it as a fall from rest).
5. With constant air resistance, the time of ascent is less than the time of descent and the return speed is less than the launch speed.

Problem Solving Hints

1. Convert km/h to m/s by multiplying by 5/18 before using g = 10 m/s² or 9.8 m/s².
2. For two bodies meeting, write both position equations from one origin and one clock, then equate positions.
3. For a body stopped and released again, restart the clock with zero velocity at that point.