Differentiation and Displacement, Velocity and Acceleration
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 over time. If the particle moves forward, its velocity is positive, and if it moves backwards, 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 concerning 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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