Velocity and Displacement (AI HL)
Once you have a velocity function \(v(t)\), integrating it recovers displacement - but you have to be careful about the difference between displacement (a net, signed quantity) and total distance travelled (always positive). This sub-topic focuses on that distinction, plus finding when a particle is at rest or moving at its fastest. It's part of the broader Kinematics topic.
13 questions on this sub-topic.
The key formulas
Covered under IB syllabus reference AHL5.13: kinematic problems involving displacement \(s\), velocity \(v=\dfrac{ds}{dt}\), with displacement recovered as \(\displaystyle\int_{t_1}^{t_2} v(t)\,dt\) and total distance travelled as \(\displaystyle\int_{t_1}^{t_2} |v(t)|\,dt\). Both are in the formula booklet.
Velocity from displacement
\(v=\dfrac{ds}{dt}\)
Differentiate \(s(t)\) to get \(v(t)\); reverse this by integrating \(v(t)\) to recover \(s(t)\).
Displacement over an interval
Integrate \(v(t)\) directly between the two times.
Regions where \(v<0\) subtract from the total - that's what makes it "net" displacement.
✓ In the formula bookletNeed the full syllabus wording and formula-booklet reference table? See Kinematics.
Worked examples
A particle moves with velocity \(v(t) = 4t - 3\) m/s.
Find the displacement between \(t = 0\) and \(t = 2\).
Worked solution
\(\int_0^2 v\,dt\) M1 \(= \int_0^2(4t - 3)\,dt.\) A1
\([2t^2 - 3t]_0^2\) M1 \(= 2\) m. A1
A particle has velocity \(v(t) = 3t^2 - 12\) m/s for \(0 \le t \le 3.\) Find the total distance travelled.
Worked solution
\(v = 0 \Rightarrow t^2 = 4 \Rightarrow t\) M1
\(= 2\) (in range). A1
\(\int_0^2(3t^2-12)\,dt\) M1
\(= -16\) A1
\(\int_2^3 = 7.\) A1
Total \(= 16 + 7 = 23\) m. A1
A particle has velocity \(v(t) = t^2 - 9\) m/s for \(t \ge 0.\)
(a) Determine when the particle moves in the negative direction.
(b) State what happens at \(t = 3.\)
Worked solution
(a) \(v < 0\) when \(t^2 < 9\), i.e. \(0 \le t < 3.\) M1 - \(v<0\) A1
(b) At \(t = 3\), \(v\) M1
\(= 0\): instantaneously at rest A1
changes direction (neg → pos). A1
A particle has velocity \(v(t) = 12t - 3t^2\) m/s for \(0 \le t \le 4.\)
(a) Find the time at which the velocity is maximum.
(b) Find the maximum velocity.
Worked solution
(a) \(a = \dfrac{dv}{dt} = 12 - 6t\) M1
\(= 0\) A1
\(\Rightarrow t = 2\) s. A1
(b) \(v(2) = 24 - 12\) M1
\(= 12\) m/s. A1
Common mistakes
- Confusing displacement with distance travelled. If velocity changes sign anywhere in the interval, \(\int v\,dt\) alone gives you displacement, not distance - forward and backward motion partly cancel out.
- Misreading a negative velocity as "stopped" rather than "moving backward". Zero velocity means momentarily at rest; a negative value means motion in the opposite direction, which still counts toward distance travelled.
- Mixing up maximum displacement with maximum speed. Displacement is greatest when \(v=0\); maximum speed happens at a different time in general, and needs its own stationary-point analysis on \(|v(t)|\) or on \(a(t)=0\).
Ready to practise properly?
13 velocity and displacement questions, marked instantly like the real exam.
Quick answers
How do you find displacement from a velocity function?
Integrate: displacement between \(t_1\) and \(t_2\) equals \(\displaystyle\int_{t_1}^{t_2} v(t)\,dt\).
What is the difference between displacement and distance travelled?
Displacement is \(\int v(t)\,dt\), so forward and backward motion cancel. Distance is \(\int |v(t)|\,dt\), so every part of the motion adds positively.