Calculus in Kinematics: Displacement, Velocity and Acceleration
Kinematics applies differentiation to motion in a straight line. If displacement s is given as a function of time t, then velocity is the rate of change of displacement, so v = ds/dt, and acceleration is the rate of change of velocity, so a = dv/dt, which is the second derivative of s. The signs carry meaning: a negative velocity means the object is moving in the negative direction, and a negative acceleration means the velocity is decreasing, which is not the same as the object slowing down unless the velocity is positive. The object is instantaneously at rest when v = 0, and that condition is what most questions in the topic turn on.
Before you start
Make sure you're comfortable with these topics first:
Method
- Identify which quantity you are given. If the question gives displacement, differentiate once for velocity and twice for acceleration; if it gives velocity, differentiate once for acceleration.
- Differentiate with respect to t using the same rule as for x: the derivative of at to the power n is ant to the power (n - 1).
- For the moment an object is instantaneously at rest, set v = 0 and solve. There is often more than one such time, and both usually matter.
- For the displacement at a particular time, substitute into the ORIGINAL displacement function, not into the velocity function.
- For maximum or minimum velocity, set the acceleration to zero and solve, because velocity is stationary where its own derivative vanishes.
- Interpret signs in context and say what they mean: a negative displacement means the object is on the negative side of the origin, a negative velocity means it is travelling backwards, and it is decelerating only when velocity and acceleration have opposite signs.
Worked example
A particle moves in a straight line so that its displacement in metres from a fixed point after t seconds is s = t cubed - 6t squared + 9t. Find (a) expressions for the velocity and acceleration, (b) the times at which the particle is instantaneously at rest, and (c) the acceleration at the later of those times.
- Differentiate s for velocity: v = ds/dt = 3t squared - 12t + 9.
- Differentiate again for acceleration: a = dv/dt = 6t - 12.
- For the particle at rest, set v = 0: 3t squared - 12t + 9 = 0, and divide by 3 to get t squared - 4t + 3 = 0.
- Factorise: (t - 1)(t - 3) = 0, so t = 1 second and t = 3 seconds.
- The later time is t = 3, so substitute into the acceleration: a = 6(3) - 12 = 6.
- Answers: v = 3t squared - 12t + 9 metres per second, a = 6t - 12 metres per second squared, the particle is at rest at t = 1 and t = 3 seconds, and at t = 3 the acceleration is 6 metres per second squared, positive so the velocity is increasing from zero.
Practice questions
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Q1How do you obtain velocity from a displacement function?Show answer
Answer: Differentiate the displacement with respect to time.
Q2s = 4t squared + 3t. Find the velocity at t = 2.Show answer
Answer: v = 8t + 3, so at t = 2 the velocity is 19.
Q3v = 5t squared - 20. Find the acceleration.Show answer
Answer: a = 10t.
Q4What condition identifies the moment a particle is instantaneously at rest?Show answer
Answer: Its velocity is zero.
Q5s = t cubed - 3t. Find the times when the particle is at rest.Show answer
Answer: v = 3t squared - 3 = 0 gives t squared = 1, so t = 1 (taking positive time).
Q6Does a negative acceleration always mean an object is slowing down?Show answer
Answer: No. It means the velocity is decreasing. If the velocity is already negative, a negative acceleration makes it more negative, so the object speeds up in the negative direction.
Q7How would you find the maximum velocity of a particle?Show answer
Answer: Set the acceleration to zero and solve for t, then substitute that time into the velocity function.
Exam-style questions
Written in the style of a IGCSE Maths exam paper, with a full mark scheme.
A particle moves so that its displacement after t seconds is s = 2t cubed - 15t squared + 24t metres. (a) Find the velocity after 4 seconds. (b) Find the times at which the particle is at rest. (c) Find the displacement at the later of those times.
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The velocity of a particle is v = t squared - 8t + 12 metres per second. Find the minimum velocity and the time at which it occurs, and explain how you know it is a minimum.
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Free printable worksheet
Want more practice on paper? Download the calculus in kinematics: displacement, velocity and acceleration worksheet pack - 7 pages of exam-style questions with a full mark scheme. One email opens every download in this browser for 14 days - no account, no card. Print it for personal and classroom use.
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