Displacement
Displacement is the change in the position of a body. Displacement is a vector quantity. Displacement has a magnitude and a direction. The magnitude is the straight-line separation between the initial position and the final position. The direction is from the initial position to the final position. Displacement depends only on these two positions. Displacement does not depend on the path taken between them.
Compare the two readings in Fig 3.1. The distance reading grows the whole time. The displacement reading grows, then shrinks when the particle turns back towards the line of A, then grows again. At every moment the displacement is smaller than the distance or equal to it. The two are equal only while the path is a straight line away from A.
The displacement of a body in a time interval is the vector from the position of the body at the start of the interval to the position of the body at the end of the interval. On a straight line, displacement is the final position minus the initial position, \(\Delta x = x_2 - x_1\), and the sign gives the direction.
On a straight line the arithmetic is simple. Let the body start at \(x_1 = +3\,\text{m}\) and end at \(x_2 = -2\,\text{m}\). The displacement is \(\Delta x = x_2 - x_1 = -2 - 3 = -5\,\text{m}\). The magnitude is \(5\,\text{m}\). The negative sign means that the displacement points in the negative direction of the axis.
A cricket ball is thrown from the bowler to the wicket-keeper along a straight line, \(20\,\text{m}\) away, and is thrown straight back to the bowler. For the whole journey, which statement is correct?
Average velocity
Average velocity is the displacement of a body divided by the time interval in which the displacement occurred. Average velocity is a vector. On a straight line, the sign of the average velocity is the sign of the displacement.
\[ \bar{v} = \dfrac{\Delta x}{\Delta t} = \dfrac{x_2 - x_1}{t_2 - t_1} \]
Here \(x_1\) is the position at time \(t_1\) and \(x_2\) is the position at the later time \(t_2\). The bar over \(v\) marks an average.
The average velocity of a body over a time interval is the displacement of the body in that interval divided by the duration of the interval. Average velocity has the direction of the displacement.
A particle starts at \(x = 0\). It moves to \(x = +5\,\text{m}\) in \(2\,\text{s}\). It then turns back and reaches \(x = +2\,\text{m}\) after another \(3\,\text{s}\). Find the average speed and the average velocity for the whole \(5\,\text{s}\).
Positions: \(x = 0\) at \(t = 0\), \(x = +5\,\text{m}\) at \(t = 2\,\text{s}\), \(x = +2\,\text{m}\) at \(t = 5\,\text{s}\). Take the positive direction to the right.
Distance \(= 5 + 3 = 8\,\text{m}\). Displacement \(= x_{\text{final}} - x_{\text{initial}} = 2 - 0 = +2\,\text{m}\).
Average speed \(= 8/5 = 1.6\,\text{m/s}\). Average velocity \(= +2/5 = +0.4\,\text{m/s}\).
The average speed is \(1.6\,\text{m/s}\). The average velocity is \(+0.4\,\text{m/s}\), directed to the right. The two differ because the particle reversed its direction.
Can the average velocity of a body over an interval be zero while the body has moved a non-zero distance in that interval?
Speed and velocity are different quantities
Speed and velocity are related, but they are not the same. Speed is a scalar built from distance. Velocity is a vector built from displacement. Because distance is never less than the magnitude of displacement, the average speed is never less than the magnitude of the average velocity. The two are equal only when the body moves along a straight line without reversing.
For any motion, the distance is \(\int |v|\,dt\) and the displacement is \(\int v\,dt\). The two integrals are equal only when \(v\) never changes sign. Every reversal of direction makes the distance larger than the magnitude of the displacement.
A runner completes exactly half of a circular track of radius \(50\,\text{m}\) in \(25\,\text{s}\). Find the magnitude of the average velocity, in m/s. Use \(\pi = 3.14\).
Why can the average velocity over an interval be zero while the average speed is not?
Average velocity is displacement divided by time, and the displacement is zero whenever the body returns to its starting point. Average speed is distance divided by time, and the distance only grows while the body moves. So a round trip has zero average velocity and a positive average speed.
A food delivery application shows a rider the distance of a trip, \(6.5\,\text{km}\), and the straight-line separation of the two addresses, \(2.1\,\text{km}\). The first number is the distance along the roads and decides the fuel and the time. The second number is the magnitude of the displacement and decides the delivery fee in some cities. The rider is paid for the displacement but rides the distance.
- Using the length of the path where the question asks for displacement.Displacement is final position minus initial position. Draw the two end points and join them with a straight arrow.
- Writing average velocity without a sign on a straight line.Decide the positive direction first. A displacement against that direction gives a negative average velocity.
- Assuming that average speed equals the magnitude of average velocity.They are equal only when the body never reverses direction. On any round trip the average velocity is zero.