Differentiation · Ordinary Level
Distance, Speed & Acceleration
Ordinary Level · Class 4 of 4 · Tap NEXT to begin
Section 1 of 1
Distance, Speed & Acceleration
If a moving object’s distance is a function of time, then speed is the rate of change of distance (the first derivative), and acceleration is the rate of change of speed (the second derivative).
The three quantities
$s = $ distance $= f(t)$ [m]. $v = $ speed $= \dfrac{ds}{dt}$ [m/s]. $a = $ acceleration $= \dfrac{d^2s}{dt^2}$ [m/s$^2$]. Initial means $t = 0$; at rest means $\dfrac{ds}{dt} = 0$.
Worked example — distance, speed, acceleration
A particle’s distance is $s = 3t^2 + 5t + 1$ (metres). Find (i) the distance after $2$ s, (ii) the speed after $3$ s, (iii) the acceleration.
(i) $s = 3(2)^2 + 5(2) + 1 = 23$ m
(ii) $\dfrac{ds}{dt} = 6t + 5$; at $t = 3$: $v = 6(3) + 5 = 23$ m/s
(iii) $a = \dfrac{d^2s}{dt^2} = 6$ m/s$^2$
$23$ m, $23$ m/s, $6$ m/s$^2$
Worked example — a cubic distance, initial speed
$s = 2t^3 + 6t^2 - 3t + 5$. Find (i) the distance after $2$ s, (ii) the initial speed, (iii) the acceleration after $10$ s.
(i) $s = 2(2)^3 + 6(2)^2 - 3(2) + 5 = 39$ m
(ii) $\dfrac{ds}{dt} = 6t^2 + 12t - 3$; initial ($t=0$): $v_0 = -3$ m/s
(iii) $a = 12t + 12$; at $t = 10$: $a = 132$ m/s$^2$
$39$ m, $-3$ m/s, $132$ m/s$^2$
You try
$s = t^2 - 6t + 20$. When is the particle at rest, and what is the distance at that time?
At rest means $\tfrac{ds}{dt} = 0$. Solve for $t$, then sub into $s$.
$\dfrac{ds}{dt} = 2t - 6 = 0 \Rightarrow t = 3$ s
$s = 3^2 - 6(3) + 20 = 11$ m
At rest at $t = 3$ s, distance $= 11$ m
That’s Differentiation.
Differentiate, then substitute — that’s the whole topic. Slope, the rule, tangents, points, max/min, increasing/decreasing, sketching, and motion.