CBSE Class 9 Physics Motion Notes Set 10

Distance and Displacement

Distance is the length of the actual path travelled and is a scalar. Displacement is the straight line from the initial to the final position, in the direction from start to finish, and is a vector.
1. Distance is never negative and never decreases; displacement can be positive, negative or zero.
2. If the distance covered is zero, the displacement is zero; but a body that returns to its starting point has zero displacement though its distance is not zero.
3. The magnitude of the displacement is never greater than the distance.

Speed, Velocity and Acceleration

1. Speed = distance / time (scalar, unit m/s). Average speed = total distance / total time.
2. Velocity = displacement / time (vector). Uniform velocity means equal displacements in equal intervals of time in a fixed direction.
3. Acceleration = change of velocity / time taken, in m/s². It is uniform when the velocity changes by equal amounts in equal intervals.
4. Retardation (deceleration) is a negative acceleration: use -a in the equations.

Equations of Uniformly Accelerated Motion

For initial velocity u, final velocity v, acceleration a, time t and displacement s:
\[ v = u + at \qquad s = ut + \tfrac12 at^2 \qquad v^2 = u^2 + 2as \]
1. The average velocity is \( \dfrac{u + v}{2} \) and s = average velocity x time.
2. Distance travelled in the n-th second: \( s_n = u + \dfrac{a}{2}(2n - 1) \).
3. For free fall replace a by g = 9.8 m/s² (taken as 10 m/s² when stated). For a body thrown upwards use -g: it rises to \( \dfrac{u^2}{2g} \) in \( \dfrac{u}{g} \) seconds and returns to the starting point with the same speed after \( \dfrac{2u}{g} \) seconds.
4. The velocity at the midpoint of the distance of a uniformly accelerated motion is \( \sqrt{\dfrac{u^2 + v^2}{2}} \), which is not the same as the average velocity.

Graphs of Motion

1. Position-time graph: for uniform velocity it is a straight line whose slope is the velocity; for variable velocity it is a curve and the velocity at an instant is the slope of the tangent.
2. Velocity-time graph: a line parallel to the time axis means uniform velocity and the area under it is the displacement. A sloping straight line means uniform acceleration (slope = acceleration), and the area of the trapezium under it gives \( s = \tfrac12 (u + v)t \).
3. If the acceleration is not uniform the velocity-time graph is a curve.
4. The area under an acceleration-time graph gives the change in velocity.

Worked Examples

Example 1. A vehicle travelling at 18 km/h stops in 10 m. Find the retardation and the time.
u = 5 m/s, v = 0: \( 0 = 25 + 2a(10) \Rightarrow a = -1.25 \) m/s². Then \( 0 = 5 - 1.25t \Rightarrow t = 4 \) s.

Example 2. A bullet is fired vertically upward at 29.4 m/s. Find the height reached and the time of flight.
\( 0 = 29.4^2 - 2(9.8)s \Rightarrow s = 44.1 \) m. Time up = 29.4 / 9.8 = 3 s, so the total time of flight is 6 s.

Example 3. A body starts from rest, accelerates at 3 m/s² for 10 s, moves at constant velocity for 20 s and then stops in 5 s. Find the displacement.
Velocity after 10 s = 30 m/s. The velocity-time graph is a trapezium whose parallel sides are 20 s (the constant-velocity part) and 35 s (the whole journey) and whose height is 30 m/s, so displacement = \( \tfrac12 (20 + 35)(30) = 825 \) m.