Variable acceleration Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Variable acceleration questions. See exactly how to solve problems on velocity, differentiation, acceleration, substitution.

velocitydifferentiationaccelerationsubstitutionat restsolving v=0
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A particle moves in a straight line so that its displacement from a fixed point is s=t2+3ts = t^{2} + 3 t metres at time tt seconds. Find its velocity vv as a function of tt.

Worked solution

  1. Write down the displacement function

    s=t2+3ts = t^{2} + 3 t

    The motion is described by this displacement-time function; velocity is its rate of change.

  2. Differentiate the displacement with respect to time

    dsdt=2t+3\frac{ds}{dt} = 2 t + 3

    Differentiating each term with the power rule gives the velocity function.

  3. State the velocity function

    v=2t+3v = 2 t + 3

    This derivative is the velocity of the particle as a function of time.

Answer
v=2t+3v = 2 t + 3
Question 2
2 markseasy
A particle moves in a straight line so that its displacement from a fixed point is s=2t3−ts = 2 t^{3} - t metres at time tt seconds. Find its velocity vv as a function of tt.

Worked solution

  1. Write down the displacement function

    s=2t3−ts = 2 t^{3} - t

    The motion is described by this displacement-time function; velocity is its rate of change.

  2. Differentiate the displacement with respect to time

    dsdt=6t2−1\frac{ds}{dt} = 6 t^{2} - 1

    Differentiating each term with the power rule gives the velocity function.

  3. State the velocity function

    v=6t2−1v = 6 t^{2} - 1

    This derivative is the velocity of the particle as a function of time.

Answer
v=6t2−1v = 6 t^{2} - 1
Question 3
2 markseasy
A particle moves in a straight line so that its displacement from a fixed point is s=t2−5t+4s = t^{2} - 5 t + 4 metres at time tt seconds. Find its velocity vv as a function of tt.

Worked solution

  1. Write down the displacement function

    s=t2−5t+4s = t^{2} - 5 t + 4

    The motion is described by this displacement-time function; velocity is its rate of change.

  2. Differentiate the displacement with respect to time

    dsdt=2t−5\frac{ds}{dt} = 2 t - 5

    Differentiating each term with the power rule gives the velocity function.

  3. State the velocity function

    v=2t−5v = 2 t - 5

    This derivative is the velocity of the particle as a function of time.

Answer
v=2t−5v = 2 t - 5
Question 4
2 markseasy
A particle moves in a straight line so that its displacement from a fixed point is s=4t3+2t2s = 4 t^{3} + 2 t^{2} metres at time tt seconds. Find its velocity vv as a function of tt.

Worked solution

  1. Write down the displacement function

    s=4t3+2t2s = 4 t^{3} + 2 t^{2}

    The motion is described by this displacement-time function; velocity is its rate of change.

  2. Differentiate the displacement with respect to time

    dsdt=12t2+4t\frac{ds}{dt} = 12 t^{2} + 4 t

    Differentiating each term with the power rule gives the velocity function.

  3. State the velocity function

    v=12t2+4tv = 12 t^{2} + 4 t

    This derivative is the velocity of the particle as a function of time.

Answer
v=12t2+4tv = 12 t^{2} + 4 t
Question 5
2 markseasy
A particle moves in a straight line with velocity v=6t−4v = 6 t - 4 (m s−1^{-1}) at time tt seconds. Find its acceleration aa as a function of tt.

Worked solution

  1. Write down the velocity function

    v=6t−4v = 6 t - 4

    Acceleration is the rate of change of velocity, so start from the velocity function.

  2. Differentiate the velocity with respect to time

    dvdt=6\frac{dv}{dt} = 6

    Differentiating the velocity with the power rule gives the acceleration.

  3. State the acceleration

    a=6a = 6

    This derivative is the acceleration of the particle.

Answer
a=6a = 6

Unlock 65 more Variable acceleration questions

Create a free account to work through every A-Level Variable acceleration question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Variable acceleration practice

Hard QuestionsPractice Questions

Related Mechanics topics

Applications of forcesConnected particlesConstant accelerationForces and frictionForces and Newton’s lawsFurther kinematics