A-Level Variable acceleration Practice Questions

Free A-Level Variable acceleration practice questions with full step-by-step worked solutions. Covers velocity, differentiation, acceleration, substitution. Practise exam-style problems and check your method.

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.
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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 has velocity v=t2−9v = t^{2}-9 (m s−1^{-1}) at time tt seconds, with t≥0t \ge 0. At which time is it instantaneously at rest?
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Worked solution

  1. Identify what the question is asking

    compare each option with the calculus definitions\text{compare each option with the calculus definitions}

    Velocity is the derivative of displacement and acceleration the derivative of velocity; use this to test each option.

  2. Set the velocity to zero

    t2−9=0⇒t=3 (t≥0)t^{2}-9=0 \Rightarrow t=3 \ (t\ge 0)

    The particle is at rest when v=0; solving t^{2}=9 with t≥0 gives t=3.

  3. State the correct choice

    correct choice: t=3 s\text{correct choice: } t=3\text{ s}

    The first option is consistent with the definitions above and is therefore correct.

Answer
t=3 st=3\text{ s}
Question 3
3 marksintermediate
A particle has displacement s=t3−3t2s = t^{3}-3t^{2} metres at time tt seconds. Which of these is its velocity v(t)v(t)?
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Worked solution

  1. Identify what the question is asking

    compare each option with the calculus definitions\text{compare each option with the calculus definitions}

    Velocity is the derivative of displacement and acceleration the derivative of velocity; use this to test each option.

  2. Differentiate the displacement

    ddt(t3−3t2)=3t2−6t\frac{d}{dt}\left(t^{3}-3t^{2}\right)=3t^{2}-6t

    Differentiating each term with the power rule gives the velocity.

  3. Rule out the second option

    option 2:inconsistent with the definitions\text{option } 2: \text{inconsistent with the definitions}

    Option 2 does not follow from differentiating displacement to velocity and velocity to acceleration, so it is rejected.

  4. Rule out the third option

    option 3:inconsistent with the definitions\text{option } 3: \text{inconsistent with the definitions}

    Option 3 does not follow from differentiating displacement to velocity and velocity to acceleration, so it is rejected.

  5. Rule out the fourth option

    option 4:inconsistent with the definitions\text{option } 4: \text{inconsistent with the definitions}

    Option 4 does not follow from differentiating displacement to velocity and velocity to acceleration, so it is rejected.

  6. State the correct choice

    correct choice: v=3t2−6t\text{correct choice: } v = 3t^{2}-6t

    The first option is consistent with the definitions above and is therefore correct.

Answer
v=3t2−6tv = 3t^{2}-6t
Question 4
5 markshard
A particle moving in a straight line has velocity v(t)v(t). What does a negative value of vv indicate?
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Worked solution

  1. Identify what the question is asking

    compare each option with the calculus definitions\text{compare each option with the calculus definitions}

    Velocity is the derivative of displacement and acceleration the derivative of velocity; use this to test each option.

  2. Interpret the sign of velocity

    v<0⇒negative directionv<0 \Rightarrow \text{negative direction}

    The sign of the velocity gives the direction of motion; a negative velocity means motion in the negative direction.

  3. Rule out the second option

    option 2:inconsistent with the definitions\text{option } 2: \text{inconsistent with the definitions}

    Option 2 does not follow from differentiating displacement to velocity and velocity to acceleration, so it is rejected.

  4. Rule out the third option

    option 3:inconsistent with the definitions\text{option } 3: \text{inconsistent with the definitions}

    Option 3 does not follow from differentiating displacement to velocity and velocity to acceleration, so it is rejected.

  5. Rule out the fourth option

    option 4:inconsistent with the definitions\text{option } 4: \text{inconsistent with the definitions}

    Option 4 does not follow from differentiating displacement to velocity and velocity to acceleration, so it is rejected.

  6. Rule out the fifth option

    option 5:inconsistent with the definitions\text{option } 5: \text{inconsistent with the definitions}

    Option 5 does not follow from differentiating displacement to velocity and velocity to acceleration, so it is rejected.

  7. Recall how velocity and acceleration are defined

    v=dsdt,a=dvdt=d2sdt2v=\frac{ds}{dt}, \quad a=\frac{dv}{dt}=\frac{d^{2}s}{dt^{2}}

    Velocity is the first derivative of displacement and acceleration is the derivative of velocity.

  8. Recall that integration reverses differentiation

    v=∫a dt,s=∫v dtv=\int a\,dt, \quad s=\int v\,dt

    Integrating acceleration gives velocity, and integrating velocity gives displacement.

  9. Recall the power rule for differentiation

    ddt(tn)=n tn−1\frac{d}{dt}\left(t^{n}\right)=n\,t^{n-1}

    Each term is differentiated by multiplying by the power and reducing the power by one.

  10. State the correct choice

    correct choice: v<0: negative direction\text{correct choice: } v<0:\ \text{negative direction}

    The first option is consistent with the definitions above and is therefore correct.

Answer
The particle is moving in the negative direction.
Question 5
8 markschallenging
Explain why solving dvdt=0\dfrac{dv}{dt}=0 locates the maximum or minimum velocity of a particle.
Show worked solution

Worked solution

  1. Identify what the question is asking

    compare each option with the calculus definitions\text{compare each option with the calculus definitions}

    Velocity is the derivative of displacement and acceleration the derivative of velocity; use this to test each option.

  2. Use the theory of stationary points

    dvdt=a=0 at a turning point of v\frac{dv}{dt}=a=0 \text{ at a turning point of } v

    A maximum or minimum of a function occurs where its derivative is zero; the derivative of velocity is the acceleration.

  3. Rule out the second option

    option 2:inconsistent with the definitions\text{option } 2: \text{inconsistent with the definitions}

    Option 2 does not follow from differentiating displacement to velocity and velocity to acceleration, so it is rejected.

  4. Rule out the third option

    option 3:inconsistent with the definitions\text{option } 3: \text{inconsistent with the definitions}

    Option 3 does not follow from differentiating displacement to velocity and velocity to acceleration, so it is rejected.

  5. Rule out the fourth option

    option 4:inconsistent with the definitions\text{option } 4: \text{inconsistent with the definitions}

    Option 4 does not follow from differentiating displacement to velocity and velocity to acceleration, so it is rejected.

  6. Rule out the fifth option

    option 5:inconsistent with the definitions\text{option } 5: \text{inconsistent with the definitions}

    Option 5 does not follow from differentiating displacement to velocity and velocity to acceleration, so it is rejected.

  7. Recall how velocity and acceleration are defined

    v=dsdt,a=dvdt=d2sdt2v=\frac{ds}{dt}, \quad a=\frac{dv}{dt}=\frac{d^{2}s}{dt^{2}}

    Velocity is the first derivative of displacement and acceleration is the derivative of velocity.

  8. Recall that integration reverses differentiation

    v=∫a dt,s=∫v dtv=\int a\,dt, \quad s=\int v\,dt

    Integrating acceleration gives velocity, and integrating velocity gives displacement.

  9. Recall the power rule for differentiation

    ddt(tn)=n tn−1\frac{d}{dt}\left(t^{n}\right)=n\,t^{n-1}

    Each term is differentiated by multiplying by the power and reducing the power by one.

  10. Recall the power rule for integration

    ∫tn dt=tn+1n+1+c\int t^{n}\,dt=\frac{t^{n+1}}{n+1}+c

    Each term is integrated by raising the power by one and dividing by the new power.

  11. Remember the constant of integration

    v(t)=∫a dt+cv(t)=\int a\,dt + c

    An indefinite integral always includes an unknown constant that an initial condition will fix.

  12. Note the units of the quantities involved

    [s]=m, [v]=m s−1, [a]=m s−2[s]=\text{m},\ [v]=\text{m s}^{-1},\ [a]=\text{m s}^{-2}

    Keeping track of units is a quick check that the calculus has been applied correctly.

  13. Recall the condition for the particle to be at rest

    v(t)=0v(t)=0

    The particle is instantaneously at rest whenever its velocity is zero.

  14. Recall the condition for maximum or minimum velocity

    a(t)=dvdt=0a(t)=\frac{dv}{dt}=0

    Velocity has a stationary value where its derivative, the acceleration, is zero.

  15. State the correct choice

    correct choice: a=dvdt=0\text{correct choice: } a=\frac{dv}{dt}=0

    The first option is consistent with the definitions above and is therefore correct.

Answer
At a maximum or minimum of vv, the rate of change of velocity is zero, and dvdt\dfrac{dv}{dt} is the acceleration aa.

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