Hard A-Level Variable acceleration Questions

Challenging, exam-style A-Level Variable acceleration questions with worked solutions. Stretch yourself on the hardest displacement, integration, initial conditions, maximum velocity problems.

displacementintegrationinitial conditionsmaximum velocitya=0distance
A-Level34 questionsStep-by-step solutions
Question 1
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.
Question 2
8 markschallenging
For v=t2−4t+3v = t^{2}-4t+3 (m s−1^{-1}) on 0≤t≤40\le t\le 4, why does the total distance travelled differ from the displacement?
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. Examine the sign of the velocity

    v=(t−1)(t−3)=0⇒t=1, t=3v=(t-1)(t-3)=0 \Rightarrow t=1,\ t=3

    Between t=1 and t=3 the velocity is negative, so the particle moves backwards; this backward travel cancels in the displacement but still counts towards the distance.

  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: t=1, t=3\text{correct choice: } t=1,\ t=3

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

Answer
Because vv changes sign at t=1t=1 and t=3t=3, so parts of the motion are in opposite directions and cancel in the displacement but not in the distance.
Question 3
8 markschallenging
A particle moves in a straight line with velocity v=3t2−9v = 3 t^{2} - 9 (m s−1^{-1}) at time tt seconds. When t=0t = 0 its displacement is 00 m. Find its displacement ss as a function of tt.
Show worked solution

Worked solution

  1. Write down the velocity function

    v=3t2−9v = 3 t^{2} - 9

    Displacement is found by integrating velocity with respect to time.

  2. Integrate the velocity

    s=∫v dt=t3−9t+cs = \int v\,dt = t^{3} - 9 t + c

    Integrating each term gives the displacement, plus a constant of integration.

  3. Use the initial condition to find the constant

    0+c=0⇒c=00 + c = 0 \Rightarrow c = 0

    Substituting the known displacement at t = 0 determines the constant c.

  4. 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.

  5. 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.

  6. 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.

  7. 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.

  8. 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.

  9. 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.

  10. 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.

  11. 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.

  12. Recall how the distance travelled is found

    distance=∫t1t2∣v∣ dt\text{distance}=\int_{t_{1}}^{t_{2}}\lvert v\rvert\,dt

    Distance uses the magnitude of velocity so that motion in either direction adds up.

  13. Recall how displacement over an interval is found

    Δs=s(t2)−s(t1)=∫t1t2v dt\Delta s = s(t_{2})-s(t_{1})=\int_{t_{1}}^{t_{2}} v\,dt

    Net displacement is the change in the displacement function between the two times.

  14. Evaluate a definite integral with its limits

    [ F(t) ]t1t2=F(t2)−F(t1)\Big[\,F(t)\,\Big]_{t_{1}}^{t_{2}}=F(t_{2})-F(t_{1})

    Substitute the upper limit, then the lower limit, and subtract.

  15. State the displacement function

    s=t3−9ts = t^{3} - 9 t

    With the constant found, this is the displacement as a function of time.

Answer
s=t3−9ts = t^{3} - 9 t
Question 4
8 markschallenging
A particle moves in a straight line with velocity v=t2−4v = t^{2} - 4 (m s−1^{-1}) at time tt seconds. The velocity changes sign in the interval [0,4][0, 4]. Find the total distance travelled between t=0t = 0 and t=4t = 4.
Show worked solution

Worked solution

  1. Write down the velocity function

    v=t2−4v = t^{2} - 4

    Because the velocity changes sign, distance uses the magnitude of the velocity.

  2. Find where the velocity changes sign

    v=0⇒t=2v = 0 \Rightarrow t = 2

    These times split the interval into parts where the velocity keeps one sign.

  3. Split the interval and take magnitudes

    distance=∫04∣v∣ dt\text{distance} = \int_{0}^{4} \lvert v\rvert\,dt

    Integrate over each sub-interval and add the magnitudes of the signed areas.

  4. Add the magnitudes of the sub-intervals

    distance=163+323=16\text{distance} = \frac{16}{3} + \frac{32}{3} = 16

    Summing the magnitudes of the signed areas gives the total distance.

  5. Integrate over the first sub-interval

    ∫02v dt=−163\int_{0}^{2} v\,dt = - \frac{16}{3}

    The signed area over this sub-interval; its magnitude contributes to the distance.

  6. Integrate over the second sub-interval

    ∫24v dt=323\int_{2}^{4} v\,dt = \frac{32}{3}

    The signed area over this sub-interval; its magnitude contributes to the distance.

  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 total distance with units

    distance=16 m\text{distance} = 16\,\text{m}

    This is the total distance travelled over the interval.

Answer
distance=16 m\text{distance} = 16\,\text{m}
Question 5
8 markschallenging
A particle moves in a straight line with displacement s=t5−t3s = t^{5} - t^{3} metres at time tt seconds. Find its acceleration when t=2t = 2.
Show worked solution

Worked solution

  1. Write down the displacement function

    s=t5−t3s = t^{5} - t^{3}

    Acceleration is the second derivative of displacement.

  2. Differentiate once to find the velocity

    dsdt=5t4−3t2\frac{ds}{dt} = 5 t^{4} - 3 t^{2}

    The first derivative of displacement is the velocity.

  3. Differentiate again to find the acceleration

    d2sdt2=20t3−6t\frac{d^{2}s}{dt^{2}} = 20 t^{3} - 6 t

    Differentiating the velocity gives the acceleration function.

  4. Substitute t = 2

    a=148a = 148

    Putting t = 2 into the acceleration function gives its value at that instant.

  5. 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.

  6. 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.

  7. 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.

  8. 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.

  9. 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.

  10. 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.

  11. 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.

  12. 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.

  13. Recall how the distance travelled is found

    distance=∫t1t2∣v∣ dt\text{distance}=\int_{t_{1}}^{t_{2}}\lvert v\rvert\,dt

    Distance uses the magnitude of velocity so that motion in either direction adds up.

  14. Recall how displacement over an interval is found

    Δs=s(t2)−s(t1)=∫t1t2v dt\Delta s = s(t_{2})-s(t_{1})=\int_{t_{1}}^{t_{2}} v\,dt

    Net displacement is the change in the displacement function between the two times.

  15. State the acceleration with units

    a=148 m s−2a = 148\,\text{m s}^{-2}

    This is the required acceleration, quoted with the correct units.

Answer
a=148 m s−2a = 148\,\text{m s}^{-2}

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