Displacement, Velocity & Acceleration
Definitions · Sign conventions · s, v, a relationships · §16.1
The calculus link between s, v, a
All three kinematic quantities are linked by differentiation and integration:
\(s\)
\(\xrightarrow{\text{diff}}\)
\(v = \dfrac{ds}{dt}\)
\(\xrightarrow{\text{diff}}\)
\(a = \dfrac{dv}{dt}\)
\(s\) \(\xleftarrow{\text{integrate}}\) \(v\) \(\xleftarrow{\text{integrate}}\) \(a\)
\(s\) \(\xleftarrow{\text{integrate}}\) \(v\) \(\xleftarrow{\text{integrate}}\) \(a\)
Sign conventions: Choose positive direction at the start. Positive displacement = in chosen direction. Negative = opposite. Velocity sign = direction of motion. Acceleration sign ≠speed up/down — it's direction of acceleration vector.
Key vocabulary
Displacement (s) — position relative to fixed origin. Can be negative.
Velocity (v) — rate of change of displacement. Negative v = moving backwards.
Speed — magnitude of velocity. Always ≥ 0. \(\text{speed}=|v|\).
Acceleration (a) — rate of change of velocity. Negative a with negative v = speeding up.
Instantaneous rest: particle is at rest when v = 0. This does NOT mean a = 0.
Using Calculus in Kinematics
Differentiating s to get v and a · Integrating a to get v and s · §16.2
A particle moves so that \(s=t^3-6t^2+9t\). Find v, a, when it is at rest, and its position then.
v = ds/dt
\(v = 3t^2-12t+9\)
a = dv/dt
\(a = 6t-12\)
v = 0
\(3t^2-12t+9=0\Rightarrow t^2-4t+3=0\Rightarrow(t-1)(t-3)=0\)
Rest at t=1 and t=3
Rest at t=1 and t=3
s at rest
\(s(1)=1-6+9=4\), \;\(s(3)=27-54+27=0\)
✓
Rest at t=1 (s=4) and t=3 (s=0)
Integrating with initial conditions
When integrating a to get v, or v to get s, you get a constant C. Use the initial condition (usually \(v=v_0\) when \(t=0\)) to find C.
\(a=6t-4\). v=8 when t=0. Find v and s (s=0 at t=0).
Integrate a
\(v=3t^2-4t+C\)
Find C
v=8 at t=0: \(C=8\Rightarrow v=3t^2-4t+8\)
Integrate v
\(s=t^3-2t^2+8t+D\)
✓
s=0 at t=0: D=0, so \(s=t^3-2t^2+8t\)
Kinematics Graphs
s-t graphs · v-t graphs · Reading and interpreting · §16.3
s-t graph (displacement-time)
Gradient = velocity
Steep upward slope → high positive velocity. Flat → at rest. Negative slope → moving backwards.
Steep upward slope → high positive velocity. Flat → at rest. Negative slope → moving backwards.
Turning point → v = 0 (particle momentarily at rest before reversing direction)
Horizontal line → particle stationary (s = constant, v = 0)
v-t graph (velocity-time)
Gradient = acceleration
Positive slope → positive acceleration. Flat → constant velocity (zero acceleration).
Positive slope → positive acceleration. Flat → constant velocity (zero acceleration).
Area under curve = displacement
Area above x-axis = positive displacement; below = negative.
Area above x-axis = positive displacement; below = negative.
Crossing x-axis → v = 0, direction reverses
Interactive — s-t and v-t for \(s=t^3-6t^2+9t\)
—
Total Distance Travelled
Distance vs displacement · Integration with direction changes · §16.4
Key distinction
Displacement = \(s(b)-s(a)\)
Net change in position. Can be positive, negative or zero. Found by \(\int_a^b v\,dt\).
Total distance = \(\int_a^b |v|\,dt\)
Always positive. Must split the integral at every t where v = 0 (changes sign), then add absolute values of each part.
- Find when v = 0 (solve \(v=0\)) between the given time limits
- Split the integral at each v=0 point
- Integrate each part and take the absolute value of each
- Add all absolute values for total distance
\(s=t^3-6t^2+9t\). Find total distance in first 4 seconds.
v = 0
\(v=3t^2-12t+9=0\Rightarrow t=1,3\)
s values
s(0)=0, s(1)=4, s(3)=0, s(4)=4
Segments
0→1: moves from 0 to 4 (+4)
1→3: moves from 4 to 0 (−4, dist=4)
3→4: moves from 0 to 4 (+4)
1→3: moves from 4 to 0 (−4, dist=4)
3→4: moves from 0 to 4 (+4)
✓
Total distance = 4+4+4 = 12
Practice Questions
3 Easy · 3 Medium · 3 Hard — Kinematics
Easy 1
[2] A particle moves so that \(s=3t^2-t\). Find its velocity and acceleration when \(t=2\).
SOLUTION
v = ds/dt
\(v=6t-1\). At \(t=2\): \(v=\boldsymbol{11}\) m/s
a = dv/dt
\(a=6\). Constant acceleration: \(\boldsymbol{6}\) m/s²
Easy 2
[2] A particle has velocity \(v=2t-6\). Find when it is momentarily at rest and which direction it moves after.
SOLUTION
At rest: v=0
\(2t-6=0\Rightarrow t=\boldsymbol{3}\) s
After t=3
\(v>0\) for \(t>3\): positive direction (e.g. rightward)
Easy 3
[3] Given \(v=\frac{ds}{dt}=4-2t\) and \(s=0\) when \(t=0\), find \(s\) as a function of \(t\).
SOLUTION
Integrate
\(s=4t-t^2+c\). At \(t=0,s=0\): \(c=0\). So \(\boldsymbol{s=4t-t^2}\)
Medium 1
[5] A particle's displacement is \(s=t^3-6t^2+9t\). Find: (a) when the particle is at rest, (b) total distance in first 4 s.
SOLUTION
v = s'
\(v=3t^2-12t+9=3(t-1)(t-3)\). At rest: \(\boldsymbol{t=1}\) and \(\boldsymbol{t=3}\)
Positions
\(s(0)=0,\; s(1)=4,\; s(3)=0,\; s(4)=4\)
Total distance
\(|4-0|+|0-4|+|4-0|=4+4+4=\boldsymbol{12}\) m
Medium 2
[4] A particle has acceleration \(a=6t-4\). When \(t=0\), \(v=3\). Find \(v\) when \(t=2\).
SOLUTION
Integrate a
\(v=3t^2-4t+c\). At \(t=0,v=3\): \(c=3\). So \(v=3t^2-4t+3\)
At t=2
\(v=12-8+3=\boldsymbol{7}\) m/s
Medium 3
[4] A particle moves with \(v=(t-2)^2\). Find the displacement from \(t=0\) to \(t=4\) and interpret the sign.
SOLUTION
Integrate
\(s=\displaystyle\int_0^4(t-2)^2\,dt=\left[\dfrac{(t-2)^3}{3}\right]_0^4=\dfrac{8}{3}-\dfrac{-8}{3}=\boldsymbol{\dfrac{16}{3}}\) m
Note
Since \(v=(t-2)^2\geq0\), the particle always moves in same direction, so total distance = displacement = \(\frac{16}{3}\) m
Hard 1
[6] Two particles A and B start from the same point at \(t=0\). A has \(v_A=3t^2-6\) and B moves with constant velocity 6 m/s. Find when they are at the same position again.
SOLUTION
s_A
\(s_A=\displaystyle\int_0^t(3u^2-6)\,du=t^3-6t\)
s_B
\(s_B=6t\)
Same position
\(t^3-6t=6t\Rightarrow t^3=12t\Rightarrow t^2=12\Rightarrow t=2\sqrt3\approx\boldsymbol{3.46}\) s (reject \(t=0\))
Hard 2
[5] A particle starts from rest and has acceleration \(a=\frac{6}{(t+1)^2}\). Find its velocity and displacement after 5 seconds.
SOLUTION
Integrate a
\(v=\displaystyle\int\dfrac{6}{(t+1)^2}\,dt=-\dfrac{6}{t+1}+c\). At \(t=0,v=0\): \(c=6\). \(v=6-\dfrac{6}{t+1}\)
Integrate v
\(s=6t-6\ln(t+1)+c\). At \(t=0,s=0\): \(c=0\).
At t=5
\(v=6-1=\boldsymbol{5}\) m/s. \(s=30-6\ln6\approx30-10.75=\boldsymbol{19.25}\) m
Hard 3
[6] A particle's velocity is \(v=e^{-t}\cos t\). Determine whether the total distance travelled for \(0\leq t<\infty\) is finite, and find it.
SOLUTION
Sign changes
\(v=0\) when \(\cos t=0\), i.e. \(t=\pi/2,3\pi/2,\ldots\) The particle oscillates with decreasing amplitude.
Convergence
Since \(|v|\leq e^{-t}\) and \(\displaystyle\int_0^\infty e^{-t}\,dt=1\), the total distance is finite.
Exact value
Using integration by parts twice: \(\displaystyle\int_0^\infty e^{-t}\cos t\,dt=\dfrac{1}{2}\). Total distance \(\leq\boldsymbol{1}\) (requires summing |areas| per interval).
Past Year Paper Questions
Cambridge 0606 — Kinematics
0606
[12]
A particle moves in a straight line. Its displacement from a fixed point \(O\) at time \(t\) seconds is \(s=(t-2)^3-3(t-2)\) metres.
(a) Find the velocity and acceleration at time \(t\). [3]
(b) Find the times when the particle is at rest and its positions then. [4]
(c) Find the total distance travelled in the first 5 seconds. [3]
(d) Sketch the \(v\)-\(t\) graph for \(0\leq t\leq5\). [2]
(a) Find the velocity and acceleration at time \(t\). [3]
(b) Find the times when the particle is at rest and its positions then. [4]
(c) Find the total distance travelled in the first 5 seconds. [3]
(d) Sketch the \(v\)-\(t\) graph for \(0\leq t\leq5\). [2]
FULL SOLUTION
(a)
Let \(u=t-2\): \(s=u^3-3u\)
\(v=\frac{ds}{dt}=3u^2-3=3(t-2)^2-3\)
\(a=\frac{dv}{dt}=6(t-2)\)
\(v=\frac{ds}{dt}=3u^2-3=3(t-2)^2-3\)
\(a=\frac{dv}{dt}=6(t-2)\)
(b)
\(v=0\Rightarrow3(t-2)^2=3\Rightarrow(t-2)^2=1\Rightarrow t=1\) or \(t=3\)
\(s(1)=(-1)^3-3(-1)=2\) m; \(s(3)=(1)^3-3(1)=-2\) m
\(s(1)=(-1)^3-3(-1)=2\) m; \(s(3)=(1)^3-3(1)=-2\) m
(c)
\(s(0)=(-2)^3-3(-2)=-2\), \(s(1)=2\), \(s(3)=-2\), \(s(5)=6\)
Distance: \(|2-(-2)|+|-2-2|+|6-(-2)|=4+4+8=\boldsymbol{16}\) m
Distance: \(|2-(-2)|+|-2-2|+|6-(-2)|=4+4+8=\boldsymbol{16}\) m
(d)
U-shaped parabola in \(v\)-\(t\), vertex at \(t=2\) where \(v=-3\). Zero crossings at \(t=1\) and \(t=3\).
0606 Style
Two-particle Problem
[8]
Particle A has displacement \(s_A = 2t^2 - 8t + 6\) and particle B has displacement \(s_B = t^2 - 2t\), both in metres.
(a) Find when the particles have the same displacement. [3]
(b) Find when the particles have the same velocity. [3]
(c) Find the acceleration of each particle at \(t = 3\). [2]
(a) Find when the particles have the same displacement. [3]
(b) Find when the particles have the same velocity. [3]
(c) Find the acceleration of each particle at \(t = 3\). [2]
FULL WORKED SOLUTION
(a)
\(2t^2-8t+6=t^2-2t\Rightarrow t^2-6t+6=0\Rightarrow t=\frac{6\pm\sqrt{12}}{2}=3\pm\sqrt{3}\)
\(\boldsymbol{t=3-\sqrt{3}\approx1.27}\) and \(\boldsymbol{t=3+\sqrt{3}\approx4.73}\)
\(\boldsymbol{t=3-\sqrt{3}\approx1.27}\) and \(\boldsymbol{t=3+\sqrt{3}\approx4.73}\)
(b)
\(v_A=4t-8\), \(v_B=2t-2\). Set equal: \(4t-8=2t-2\Rightarrow 2t=6\Rightarrow\boldsymbol{t=3}\)
(c)
\(a_A=4\text{ m/s}^2\) (constant). \(a_B=2\text{ m/s}^2\) (constant, independent of \(t\)).
0606 Style
Non-polynomial Kinematics
[7]
A particle moves with velocity \(v = \dfrac{6}{(t+1)^2}\) m/s for \(t \geq 0\).
(a) Find the acceleration at \(t = 2\). [3]
(b) Find the displacement in the first 5 seconds. [3]
(c) As \(t \to \infty\), what happens to the velocity? Explain. [1]
(a) Find the acceleration at \(t = 2\). [3]
(b) Find the displacement in the first 5 seconds. [3]
(c) As \(t \to \infty\), what happens to the velocity? Explain. [1]
FULL WORKED SOLUTION
(a)
\(a=\frac{dv}{dt}=-\frac{12}{(t+1)^3}\). At \(t=2\): \(a=-\frac{12}{27}=-\boldsymbol{\frac{4}{9}}\text{ m/s}^2\)
(b)
\(s=\int_0^5\frac{6}{(t+1)^2}dt=\left[-\frac{6}{t+1}\right]_0^5=-1+6=\boldsymbol{5}\text{ m}\)
(c)
As \(t\to\infty\), \((t+1)^2\to\infty\), so \(v\to0\). The particle decelerates and approaches rest.