Short Notes: Motion in Straight Line

2.1 Kinematic Equations for Uniform Acceleration

For motion in a straight line under constant acceleration (also called uniform acceleration), the following equations relate initial velocity, final velocity, acceleration, time and displacement. These are valid when acceleration a is constant.

Derivations (brief)

Start from definitions and constant acceleration assumption.

Velocity as a function of time:

\(v = u + at\)

Displacement as a function of time (integrate velocity):

\(s = \int_{0}^{t} (u + at)\,dt = ut + \tfrac{1}{2}at^{2}\)

Eliminate time between \(v = u + at\) and \(s = ut + \tfrac{1}{2}at^{2}\) to obtain:

\(v^{2} = u^{2} + 2as\)

2.2 Important Definitions

Remarks on sign and vectors

• Choose a sign convention (for example, take right/upwards as positive). Signs of velocity and acceleration depend on that choice.
• When velocity and acceleration have the same sign, speed increases; if signs are opposite, speed decreases.
• All kinematic equations above apply separately to each straight-line direction; treat motion as one-dimensional.

2.3 Free Fall (motion under gravity)

Free fall refers to motion under the influence of gravity alone, with no air resistance. Near Earth's surface, gravitational acceleration is approximately \(g = 9.8\ \text{m s}^{-2}\) directed downward.

• Acceleration due to gravity: \(\;a = -g\) if upward is taken as positive; \(\;a = +g\) if downward is positive.
• Motion thrown vertically upward with initial speed \(u\):
  • Time to reach maximum height: \(\;t_{\text{up}} = \dfrac{u}{g}\)
  • Maximum height above launch point: \(\;H = \dfrac{u^{2}}{2g}\)
  • Time of flight (up and down to same level): \(\;T = \dfrac{2u}{g}\)
  • Velocity on return to starting point (neglecting air resistance): \(\;v = -u\) (opposite direction, same magnitude)
• Object dropped from rest: initial velocity \(u=0\); velocity after time \(t\) is \(\;v = -gt\) (taking downward negative if upward positive), and displacement in time \(t\) is \(\;s = -\tfrac{1}{2}gt^{2}\).
• Object projected downward with initial speed \(u\): use the same kinematic equations with \(a = -g\).

Short derivation for maximum height

Use \(v^{2} = u^{2} + 2as\). At maximum height, final velocity \(v=0\) and acceleration \(a=-g\).

\(0 = u^{2} - 2gH\)

\(H = \dfrac{u^{2}}{2g}\)

Illustrative example

Example: A ball is thrown vertically upward with speed \(u = 19.6\ \text{m s}^{-1}\). Find the time to reach the top and maximum height (take \(g = 9.8\ \text{m s}^{-2}\)).

Time to reach top:

\(t_{\text{up}} = \dfrac{u}{g} = \dfrac{19.6}{9.8} = 2\ \text{s}\)

Maximum height:

\(H = \dfrac{u^{2}}{2g} = \dfrac{(19.6)^{2}}{2\times 9.8} = \dfrac{384.16}{19.6} = 19.6\ \text{m}\)

Additional useful points and common formulas

• To find displacement when acceleration changes piecewise, treat each interval with constant acceleration separately and add displacements vectorially.
• When acceleration is zero (\(a=0\)), motion is uniform: \(\;v = u =\) constant and \(\;s = ut\).
• Average velocity for uniformly accelerated motion equals the arithmetic mean of initial and final velocities: \(\;v_{\text{avg}} = \dfrac{u+v}{2}\).
• Displacement in the first second, second second, etc., can be obtained using \(s_{n} = s(n) - s(n-1)\) where \(s(t)=ut+\tfrac{1}{2}at^{2}\).

Summary

For straight-line motion with constant acceleration, memorise the three basic equations \(v = u + at\), \(s = ut + \tfrac{1}{2}at^{2}\) and \(v^{2} = u^{2} + 2as\), understand the meaning of displacement, velocity and acceleration, and apply a clear sign convention. For vertical motion under gravity, replace \(a\) by \(-g\) or \(+g\) according to the chosen positive direction and use the same equations.