## Distance

Distance is the magnitude of the total movement from the start point or a fixed point.

## Displacement

The displacement of a moving position relative to a fixed point. Displacement gives both the distance and direction that a particle is from a fixed point.

For example, a particle moves \( 5 \) units forwards from \( O \), and moves \( 3 \) backwards. Its displacement is \( 2 \) units from \( O \).

Note that distance travelled is not always necessarily the same as displacement, as displacement refers to the current position from \( O \) no matter what the particle travelled,

## Velocity

The average velocity of a particle is the rate of change of its position with respect to time. If the particle moves forwards, then its velocity is positive, and if the particle moves backwards, then its velocity is negative.

Velocity is obtained by differentiating its displacement, \( x \) in terms of \( t \).

\( v = \displaystyle \dfrac{dx}{dt} \) or \( \displaystyle x = \int v dt \)

## Speed

Instantaneous speed is the magnitude of instantaneous velocity and is always positive, regardless of its direction either forwards or backwards.

## At Rest

The particle is at rest when \( v = 0 \).

## Acceleration

The average acceleration of a particle during a time interval is the rate of change of its velocity with respect to time.

Acceleration is obtained by differentiating its velocity, \( v \) in terms of \( t \).

\( \displaystyle a = \displaystyle \dfrac{dv}{dt} \) or \( v = \displaystyle \int a dt \)

## Example 1

\( x = 3t^2 +2t + 8 \)

(a) Find the velocity in terms of \( t \).

\( \begin{align} \displaystyle v &= \dfrac{dx}{dt} \\ &= \dfrac{d}{dt} (3t^2+2t+8) \\ &= 6t+2 \end{align} \)

(b) Find the initial velocity.

\( v(0) = 6 \times 0 + 2 = 0 \)

(c) Find the initial direction.

moves forwards

## Example 2

\( v = t^2 + 4t -5 \)

(a) Find the initial velocity.

\( v(0) = 0^2 + 4 \times 4 -5 \)

(b) Find its initial direction of motion.

moves backwards

(c) Find the equation of the acceleration.

\( \displaystyle \begin{align} a &= \dfrac{dv}{dt} \\ &= \dfrac{d}{dt}(t^2+4t-5) \\ &= 2t+4 \end{align} \)

(d) Find the equation of the displacement given its initial displacement is \( 3 \).

\( \displaystyle \begin{align} x &= \int v dt \\ &= \int(t^2+4t-5)dt \\ &= \dfrac{1}{3}t^3 +2t^2 -5t + C \end{align} \)

\( t=0 \rightarrow x=3 \)

\( \displaystyle \begin{align} 3 &= \dfrac{1}{3} \times 0^3 + 2 \times 0^2 -5 \times 0 + C \\ C &= 3 \\ \therefore x &= \dfrac{1}{3}t^3 +2t^2 -5t + 3 \end{align} \)

## Example 3

\( x = t^2 – 4t + 3 \), where \( t \) in seconds.

(a) Find its velocity in terms of \( t \).

\( \displaystyle \begin{align} v &= \dfrac{dx}{dt} \\ &= \dfrac{d}{dt}(t^2-4t+3) \\ &= 2t-4 \end{align} \)

(b) Find the time when the particle is at rest.

\( \begin{align} v &= 0 \\ 2t-4 &= 0 \\ t &= 2 \end{align} \)

(c) Find the displacement when the particle is at rest.

\( \begin{align} x &= 2^2 – 4 \times 2 + 3 \\ &= -1 \end{align} \)

(d) Find the total distance travelled in the first \( 3 \) seconds.

\( \displaystyle \begin{align} x (0) &= 0^2 – 4 \times 0 + 3 = 3 \\ x (2) &= 2^2 – 4 \times 2 + 3 = -1 \\ x (3) &= 3^2 – 4 \times 3 + 3 = 0 \end{align} \)

\( \displaystyle \begin{array}{ccc} \hline \\ t & x & \text{distance} \\ \hline 0 \rightarrow 2 & 3 \rightarrow -1 & \left|3 – -1\right|=4 \\ \hline 2 \rightarrow 3 & -1 \rightarrow 0 & \left|0 – -1\right|=1 \\ \hline \end{array} \)

Therefore the total distance travelled is \( 4+1 = 5 \) units.

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