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Calculus - Kinematics (Displacement, Velocity, Acceleration)

Grade 12IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Displacement (ss): The position of an object relative to a fixed origin at time tt.

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Velocity (vv): The rate of change of displacement with respect to time. It is the derivative of ss.

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Acceleration (aa): The rate of change of velocity with respect to time. It is the derivative of vv or the second derivative of ss.

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Instantaneous Rest: Occurs when the velocity of the particle is zero (v=0v = 0).

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Initial Conditions: Values at t=0t = 0 (e.g., initial velocity or initial position), used to find the constant of integration CC.

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Turning Points: When a particle changes direction, v=0v=0 and the sign of vv changes.

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Total Distance Traveled: The sum of the absolute changes in displacement, calculated by integrating the absolute value of velocity ∫∣v∣dt\int |v| dt or checking segments between turning points.

📐Formulae

v=dsdtv = \frac{ds}{dt}

a=dvdt=d2sdt2a = \frac{dv}{dt} = \frac{d^2s}{dt^2}

s=∫v dts = \int v \, dt

v=∫a dtv = \int a \, dt

Average Velocity = Total DisplacementTotal Time\frac{\text{Total Displacement}}{\text{Total Time}}

Average Speed = Total DistanceTotal Time\frac{\text{Total Distance}}{\text{Total Time}}

💡Examples

Problem 1:

A particle moves in a straight line such that its displacement ss meters from a fixed point OO at time tt seconds is given by s=t3−6t2+9ts = t^3 - 6t^2 + 9t. Find the acceleration of the particle when it is at instantaneous rest.

Solution:

  1. Find velocity: v=dsdt=3t2−12t+9v = \frac{ds}{dt} = 3t^2 - 12t + 9.
  2. Set v=0v = 0 for rest: 3(t2−4t+3)=0  ⟹  3(t−1)(t−3)=03(t^2 - 4t + 3) = 0 \implies 3(t-1)(t-3) = 0. So, t=1t = 1 or t=3t = 3.
  3. Find acceleration: a=dvdt=6t−12a = \frac{dv}{dt} = 6t - 12.
  4. At t=1t=1: a=6(1)−12=−6 m/s2a = 6(1) - 12 = -6 \text{ m/s}^2.
  5. At t=3t=3: a=6(3)−12=6 m/s2a = 6(3) - 12 = 6 \text{ m/s}^2.

Explanation:

To find the acceleration at rest, we first differentiate displacement to get velocity, solve for tt when velocity is zero, then differentiate velocity to get the acceleration function and substitute the time values.

Problem 2:

A particle starts from rest at the origin. Its acceleration is given by a=6t−4a = 6t - 4. Find the expression for displacement ss in terms of tt.

Solution:

  1. Find velocity: v=∫(6t−4) dt=3t2−4t+Cv = \int (6t - 4) \, dt = 3t^2 - 4t + C.
  2. Use initial conditions (t=0,v=0t=0, v=0): 0=3(0)2−4(0)+C  ⟹  C=00 = 3(0)^2 - 4(0) + C \implies C = 0. So v=3t2−4tv = 3t^2 - 4t.
  3. Find displacement: s=∫(3t2−4t) dt=t3−2t2+Ks = \int (3t^2 - 4t) \, dt = t^3 - 2t^2 + K.
  4. Use initial conditions (t=0,s=0t=0, s=0): 0=03−2(0)2+K  ⟹  K=00 = 0^3 - 2(0)^2 + K \implies K = 0.
  5. Result: s=t3−2t2s = t^3 - 2t^2.

Explanation:

Integration is used to move from acceleration to velocity, and then from velocity to displacement. The constants of integration are determined using the 'start from rest' (v=0v=0) and 'at the origin' (s=0s=0) conditions.