Newton’s Equations of Motion
These equations describe motion under constant acceleration.
They relate displacement $s$, initial velocity $u$, final velocity $v$, acceleration $a$, and time $t$.
\[
v = u + at
\]
\[
s = ut + \tfrac{1}{2}at^2
\]
\[
v^2 = u^2 + 2as
\]
Deriving the equations of motion
Rearranging the Equations
From the first equation:
\[
v = u + at
\quad\Rightarrow\quad
t = \frac{v - u}{a}
\]
From the second equation:
\[
s = ut + \tfrac{1}{2}at^2
\]
\[
\Rightarrow\;
at^2 + 2ut - 2s = 0
\]
\[
\Rightarrow\;
t = \frac{-u \pm \sqrt{u^2 + 2as}}{a}
\]
From the third equation:
\[
v^2 = u^2 + 2as
\quad\Rightarrow\quad
s = \frac{v^2 - u^2}{2a}
\]
Special Case: Starting From Rest
If $u = 0$:
\[
v = at
\]
\[
s = \tfrac{1}{2}at^2
\]
\[
v^2 = 2as
\]
Rearranging for time:
\[
t = \sqrt{\frac{2s}{a}}
\]
Rearranging for acceleration:
\[
a = \frac{v^2}{2s}
\]
Displacement $s$ is the vector quantity describing the distance and direction from a fixed point.
After time $t$, the displacement from the origin can be written as a function $s(t)$.
A particle moving in a plane with coordinates $(x(t), y(t))$ at time $t$ has position vector
\[
\vec{s}(t) = x(t)\,\mathbf{i} + y(t)\,\mathbf{j}.
\]
The distance from the origin is the magnitude of the displacement:
\[
|\vec{s}(t)| = \sqrt{x(t)^2 + y(t)^2}.
\]
Velocity is the rate of change of displacement with respect to time:
\[
\vec{v}(t) = \frac{d\vec{s}}{dt}
= \frac{dx}{dt}\,\mathbf{i} + \frac{dy}{dt}\,\mathbf{j}.
\]
This is often written as:
\[
\vec{v}(t) = x'(t)\,\mathbf{i} + y'(t)\,\mathbf{j}.
\]
Or
\[
v
= \dot{x}\,\mathbf{i}
+ \dot{y}\,\mathbf{j}
\]
The speed of the particle is the magnitude of the velocity:
\[
|\vec{v}(t)| = \sqrt{\,x'(t)^2 + y'(t)^2\,}.
\]
The direction of motion is given by the angle $\theta$ where
\[
\tan\theta = \frac{y'(t)}{x'(t)}.
\]
Acceleration is the rate of change of velocity with respect to time:
\[
\vec{a}(t) = \frac{d\vec{v}}{dt}
= x''(t)\,\mathbf{i} + y''(t)\,\mathbf{j}.
\]
This is often written as:
\[
\vec{a}(t) = x''(t)\,\mathbf{i} + y''(t)\,\mathbf{j}.
\]
Or
\[
a
= \ddot{x}\,\mathbf{i}
+ \ddot{y}\,\mathbf{j}
\]
The magnitude of acceleration is:
\[
|\vec{a}(t)| = \sqrt{\,x''(t)^2 + y''(t)^2\,}.
\]
The direction of acceleration is given by:
\[
\tan\theta = \frac{y''(t)}{x''(t)}.
\]
A particle moves in a plane with displacement given by:
\[
x(t) = 3t^3 + 2t^2,
\qquad
y(t) = 4t^2 + 5t.
\]
(Distances in metres, time in seconds.)
Find, when $t = 2$:
- the position of the particle
- the magnitude and direction of its velocity
- the magnitude and direction of its acceleration
1. Position at $t = 2$
\[
x(2) = 3(2)^3 + 2(2)^2 = 24 + 8 = 32
\]
\[
y(2) = 4(2)^2 + 5(2) = 16 + 10 = 26
\]
The particle is at the point $(32, 26)$.
2. Velocity at $t = 2$
Differentiate:
\[
x'(t) = 9t^2 + 4t,
\qquad
y'(t) = 8t + 5.
\]
Evaluate at $t = 2$:
\[
x'(2) = 9(4) + 8 = 44,
\qquad
y'(2) = 16 + 5 = 21.
\]
Velocity vector:
\[
\vec{v}(2) = 44\,\mathbf{i} + 21\,\mathbf{j}.
\]
Speed:
\[
|\vec{v}(2)| = \sqrt{44^2 + 21^2}
= \sqrt{1936 + 441}
= \sqrt{2377}
\approx 48.8\text{ m/s}.
\]
Direction:
\[
\theta = \tan^{-1}\left(\frac{21}{44}\right)
\approx 25.5^\circ.
\]
The velocity is 48.8 m/s at a direction
of 25.5° from the horizontal.
3. Acceleration at $t = 2$
Differentiate again:
\[
x''(t) = 18t + 4,
\qquad
y''(t) = 8.
\]
Evaluate at $t = 2$:
\[
x''(2) = 36 + 4 = 40,
\qquad
y''(2) = 8.
\]
Acceleration vector:
\[
\vec{a}(2) = 40\,\mathbf{i} + 8\,\mathbf{j}.
\]
Magnitude:
\[
|\vec{a}(2)| = \sqrt{40^2 + 8^2}
= \sqrt{1600 + 64}
= \sqrt{1664}
\approx 40.8\text{ m/s}^2.
\]
Direction:
\[
\theta = \tan^{-1}\left(\frac{8}{40}\right)
\approx 11.3^\circ.
\]
