Hard A-Level Further kinematics Questions

Challenging, exam-style A-Level Further kinematics questions with worked solutions. Stretch yourself on the hardest position, integration, initial conditions, velocity problems.

positionintegrationinitial conditionsvelocityaccelerationsecond derivative
A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
A particle has acceleration a=6i+4tj\mathbf{a} = 6\mathbf{i} + 4t\mathbf{j} m s−2^{-2} and velocity 2i+3j2\mathbf{i} + 3\mathbf{j} m s−1^{-1} at t=0t = 0. Which is its velocity v\mathbf{v}?
Show worked solution

Worked solution

  1. Identify what the question is asking

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

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

  2. Integrate each component and apply the initial condition

    vx=6t+2, vy=2t2+3v_x=6t+2,\ v_y=2t^{2}+3

    Integrating gives 6t+c16t+c_1 and 2t2+c22t^2+c_2; at t=0t=0, c1=2c_1=2 and c2=3c_2=3.

  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 the vector kinematics definitions, 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 the vector kinematics definitions, 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 the vector kinematics definitions, 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 the vector kinematics definitions, so it is rejected.

  7. Recall the vector kinematics definitions

    v=drdt,a=dvdt\mathbf{v}=\frac{d\mathbf{r}}{dt},\quad \mathbf{a}=\frac{d\mathbf{v}}{dt}

    Velocity is the derivative of position and acceleration is the derivative of velocity.

  8. Recall that integration reverses differentiation

    v=∫a dt,r=∫v dt\mathbf{v}=\int \mathbf{a}\,dt,\quad \mathbf{r}=\int \mathbf{v}\,dt

    Integrating acceleration gives velocity, and integrating velocity gives position.

  9. Differentiate each component separately

    ddt(f(t)i+g(t)j)=f′(t)i+g′(t)j\frac{d}{dt}\bigl(f(t)\mathbf{i}+g(t)\mathbf{j}\bigr)=f'(t)\mathbf{i}+g'(t)\mathbf{j}

    With i and j fixed, differentiate the scalar coefficient of each unit vector.

  10. Integrate each component separately

    ∫(f(t)i+g(t)j) dt=(∫f dt)i+(∫g dt)j\int\bigl(f(t)\mathbf{i}+g(t)\mathbf{j}\bigr)\,dt=\Bigl(\int f\,dt\Bigr)\mathbf{i}+\Bigl(\int g\,dt\Bigr)\mathbf{j}

    Each component is integrated independently, each with its own constant.

  11. Recall how speed is defined

    speed=∣v∣=vx2+vy2\text{speed}=\lvert\mathbf{v}\rvert=\sqrt{v_x^{2}+v_y^{2}}

    Speed is the magnitude of the velocity vector, not a component.

  12. Recall how bearing is measured

    bearing is measured clockwise from north\text{bearing is measured clockwise from north}

    In navigation, bearing 000∘000^\circ is due north and 090∘090^\circ is due east.

  13. Use the east and north components for bearing

    bearing=tan⁡−1 ⁣(vxvy) adjusted for quadrant\text{bearing}=\tan^{-1}\!\left(\dfrac{v_x}{v_y}\right)\text{ adjusted for quadrant}

    With i\mathbf{i} east and j\mathbf{j} north, the bearing follows from the velocity components.

  14. Remember constants of integration in each component

    v(t)=(∫ax dt+c1)i+(∫ay dt+c2)j\mathbf{v}(t)=\Bigl(\int a_x\,dt + c_1\Bigr)\mathbf{i}+\Bigl(\int a_y\,dt + c_2\Bigr)\mathbf{j}

    Each component of an indefinite integral needs its own constant, fixed by initial conditions.

  15. State the correct choice

    correct choice: v=(6t+2)i+(2t2+3)j\text{correct choice: } \mathbf{v}=(6t+2)\mathbf{i}+(2t^{2}+3)\mathbf{j}

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

Answer
v=(6t+2)i+(2t2+3)j\mathbf{v} = (6t+2)\mathbf{i} + (2t^{2}+3)\mathbf{j}
Question 2
8 markschallenging
Why does finding displacement between t1t_1 and t2t_2 require integrating the velocity vector rather than subtracting position vectors directly from an unknown r(t)\mathbf{r}(t)?
Show worked solution

Worked solution

  1. Identify what the question is asking

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

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

  2. Use the fundamental theorem for vector components

    Δr=∫t1t2v dt\Delta\mathbf{r}=\int_{t_1}^{t_2}\mathbf{v}\,dt

    Integrating velocity over an interval gives the net change in each position component.

  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 the vector kinematics definitions, 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 the vector kinematics definitions, 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 the vector kinematics definitions, 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 the vector kinematics definitions, so it is rejected.

  7. Recall the vector kinematics definitions

    v=drdt,a=dvdt\mathbf{v}=\frac{d\mathbf{r}}{dt},\quad \mathbf{a}=\frac{d\mathbf{v}}{dt}

    Velocity is the derivative of position and acceleration is the derivative of velocity.

  8. Recall that integration reverses differentiation

    v=∫a dt,r=∫v dt\mathbf{v}=\int \mathbf{a}\,dt,\quad \mathbf{r}=\int \mathbf{v}\,dt

    Integrating acceleration gives velocity, and integrating velocity gives position.

  9. Differentiate each component separately

    ddt(f(t)i+g(t)j)=f′(t)i+g′(t)j\frac{d}{dt}\bigl(f(t)\mathbf{i}+g(t)\mathbf{j}\bigr)=f'(t)\mathbf{i}+g'(t)\mathbf{j}

    With i and j fixed, differentiate the scalar coefficient of each unit vector.

  10. Integrate each component separately

    ∫(f(t)i+g(t)j) dt=(∫f dt)i+(∫g dt)j\int\bigl(f(t)\mathbf{i}+g(t)\mathbf{j}\bigr)\,dt=\Bigl(\int f\,dt\Bigr)\mathbf{i}+\Bigl(\int g\,dt\Bigr)\mathbf{j}

    Each component is integrated independently, each with its own constant.

  11. Recall how speed is defined

    speed=∣v∣=vx2+vy2\text{speed}=\lvert\mathbf{v}\rvert=\sqrt{v_x^{2}+v_y^{2}}

    Speed is the magnitude of the velocity vector, not a component.

  12. Recall how bearing is measured

    bearing is measured clockwise from north\text{bearing is measured clockwise from north}

    In navigation, bearing 000∘000^\circ is due north and 090∘090^\circ is due east.

  13. Use the east and north components for bearing

    bearing=tan⁡−1 ⁣(vxvy) adjusted for quadrant\text{bearing}=\tan^{-1}\!\left(\dfrac{v_x}{v_y}\right)\text{ adjusted for quadrant}

    With i\mathbf{i} east and j\mathbf{j} north, the bearing follows from the velocity components.

  14. Remember constants of integration in each component

    v(t)=(∫ax dt+c1)i+(∫ay dt+c2)j\mathbf{v}(t)=\Bigl(\int a_x\,dt + c_1\Bigr)\mathbf{i}+\Bigl(\int a_y\,dt + c_2\Bigr)\mathbf{j}

    Each component of an indefinite integral needs its own constant, fixed by initial conditions.

  15. State the correct choice

    correct choice: Δr=∫v dt\text{correct choice: } \Delta\mathbf{r}=\int\mathbf{v}\,dt

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

Answer
Integration of v\mathbf{v} gives the change in each component of position; r(t2)−r(t1)=∫t1t2v dt\mathbf{r}(t_2)-\mathbf{r}(t_1)=\int_{t_1}^{t_2}\mathbf{v}\,dt when r\mathbf{r} is found from v\mathbf{v}.
Question 3
8 markschallenging
A particle moves in a plane with velocity v=12t2−4i+6t−2j\mathbf{v} = 12 t^{2} - 4\mathbf{i}+6 t - 2\mathbf{j} (m s−1^{-1}) at time tt seconds. Find its acceleration vector a\mathbf{a} as a function of tt.
Show worked solution

Worked solution

  1. Write down the velocity vector

    v=12t2−4i+6t−2j\mathbf{v}=12 t^{2} - 4\mathbf{i}+6 t - 2\mathbf{j}

    Acceleration is the rate of change of velocity.

  2. Differentiate the i\mathbf{i} component

    ddt(12t2−4)=24t\frac{d}{dt}\bigl(12 t^{2} - 4\bigr)=24 t

    Differentiate the i\mathbf{i} component of velocity.

  3. Differentiate the j\mathbf{j} component

    ddt(6t−2)=6\frac{d}{dt}\bigl(6 t - 2\bigr)=6

    Differentiate the j\mathbf{j} component of velocity.

  4. Recall the vector kinematics definitions

    v=drdt,a=dvdt\mathbf{v}=\frac{d\mathbf{r}}{dt},\quad \mathbf{a}=\frac{d\mathbf{v}}{dt}

    Velocity is the derivative of position and acceleration is the derivative of velocity.

  5. Recall that integration reverses differentiation

    v=∫a dt,r=∫v dt\mathbf{v}=\int \mathbf{a}\,dt,\quad \mathbf{r}=\int \mathbf{v}\,dt

    Integrating acceleration gives velocity, and integrating velocity gives position.

  6. Differentiate each component separately

    ddt(f(t)i+g(t)j)=f′(t)i+g′(t)j\frac{d}{dt}\bigl(f(t)\mathbf{i}+g(t)\mathbf{j}\bigr)=f'(t)\mathbf{i}+g'(t)\mathbf{j}

    With i and j fixed, differentiate the scalar coefficient of each unit vector.

  7. Integrate each component separately

    ∫(f(t)i+g(t)j) dt=(∫f dt)i+(∫g dt)j\int\bigl(f(t)\mathbf{i}+g(t)\mathbf{j}\bigr)\,dt=\Bigl(\int f\,dt\Bigr)\mathbf{i}+\Bigl(\int g\,dt\Bigr)\mathbf{j}

    Each component is integrated independently, each with its own constant.

  8. Recall how speed is defined

    speed=∣v∣=vx2+vy2\text{speed}=\lvert\mathbf{v}\rvert=\sqrt{v_x^{2}+v_y^{2}}

    Speed is the magnitude of the velocity vector, not a component.

  9. Recall how bearing is measured

    bearing is measured clockwise from north\text{bearing is measured clockwise from north}

    In navigation, bearing 000∘000^\circ is due north and 090∘090^\circ is due east.

  10. Use the east and north components for bearing

    bearing=tan⁡−1 ⁣(vxvy) adjusted for quadrant\text{bearing}=\tan^{-1}\!\left(\dfrac{v_x}{v_y}\right)\text{ adjusted for quadrant}

    With i\mathbf{i} east and j\mathbf{j} north, the bearing follows from the velocity components.

  11. Remember constants of integration in each component

    v(t)=(∫ax dt+c1)i+(∫ay dt+c2)j\mathbf{v}(t)=\Bigl(\int a_x\,dt + c_1\Bigr)\mathbf{i}+\Bigl(\int a_y\,dt + c_2\Bigr)\mathbf{j}

    Each component of an indefinite integral needs its own constant, fixed by initial conditions.

  12. Apply initial conditions to each component

    substitute t=t0 into each component separately\text{substitute } t=t_0 \text{ into each component separately}

    Initial velocity or position fixes the constants in the corresponding components.

  13. Recall the power rule for differentiation

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

    Each polynomial term is differentiated by reducing the power by one.

  14. Recall the power rule for integration

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

    Each polynomial term is integrated by raising the power by one.

  15. State the acceleration vector

    a=24ti+6j\mathbf{a}=24 t\mathbf{i}+6\mathbf{j}

    This is the acceleration as a function of time.

Answer
a=24ti+6j\mathbf{a}=24 t\mathbf{i}+6\mathbf{j}
Question 4
8 markschallenging
A particle moves in a plane so that its position vector at time tt seconds is r=2t3−3ti+t4−2tj\mathbf{r} = 2 t^{3} - 3 t\mathbf{i}+t^{4} - 2 t\mathbf{j} metres. Find its velocity vector v\mathbf{v} as a function of tt.
Show worked solution

Worked solution

  1. Write down the position vector

    r=2t3−3ti+t4−2tj\mathbf{r}=2 t^{3} - 3 t\mathbf{i}+t^{4} - 2 t\mathbf{j}

    Velocity is found by differentiating the position vector with respect to time.

  2. Differentiate the i\mathbf{i} component

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

    Apply the power rule to the i\mathbf{i} component.

  3. Differentiate the j\mathbf{j} component

    ddt(t4−2t)=4t3−2\frac{d}{dt}\bigl(t^{4} - 2 t\bigr)=4 t^{3} - 2

    Apply the power rule to the j\mathbf{j} component.

  4. Recall the vector kinematics definitions

    v=drdt,a=dvdt\mathbf{v}=\frac{d\mathbf{r}}{dt},\quad \mathbf{a}=\frac{d\mathbf{v}}{dt}

    Velocity is the derivative of position and acceleration is the derivative of velocity.

  5. Recall that integration reverses differentiation

    v=∫a dt,r=∫v dt\mathbf{v}=\int \mathbf{a}\,dt,\quad \mathbf{r}=\int \mathbf{v}\,dt

    Integrating acceleration gives velocity, and integrating velocity gives position.

  6. Differentiate each component separately

    ddt(f(t)i+g(t)j)=f′(t)i+g′(t)j\frac{d}{dt}\bigl(f(t)\mathbf{i}+g(t)\mathbf{j}\bigr)=f'(t)\mathbf{i}+g'(t)\mathbf{j}

    With i and j fixed, differentiate the scalar coefficient of each unit vector.

  7. Integrate each component separately

    ∫(f(t)i+g(t)j) dt=(∫f dt)i+(∫g dt)j\int\bigl(f(t)\mathbf{i}+g(t)\mathbf{j}\bigr)\,dt=\Bigl(\int f\,dt\Bigr)\mathbf{i}+\Bigl(\int g\,dt\Bigr)\mathbf{j}

    Each component is integrated independently, each with its own constant.

  8. Recall how speed is defined

    speed=∣v∣=vx2+vy2\text{speed}=\lvert\mathbf{v}\rvert=\sqrt{v_x^{2}+v_y^{2}}

    Speed is the magnitude of the velocity vector, not a component.

  9. Recall how bearing is measured

    bearing is measured clockwise from north\text{bearing is measured clockwise from north}

    In navigation, bearing 000∘000^\circ is due north and 090∘090^\circ is due east.

  10. Use the east and north components for bearing

    bearing=tan⁡−1 ⁣(vxvy) adjusted for quadrant\text{bearing}=\tan^{-1}\!\left(\dfrac{v_x}{v_y}\right)\text{ adjusted for quadrant}

    With i\mathbf{i} east and j\mathbf{j} north, the bearing follows from the velocity components.

  11. Remember constants of integration in each component

    v(t)=(∫ax dt+c1)i+(∫ay dt+c2)j\mathbf{v}(t)=\Bigl(\int a_x\,dt + c_1\Bigr)\mathbf{i}+\Bigl(\int a_y\,dt + c_2\Bigr)\mathbf{j}

    Each component of an indefinite integral needs its own constant, fixed by initial conditions.

  12. Apply initial conditions to each component

    substitute t=t0 into each component separately\text{substitute } t=t_0 \text{ into each component separately}

    Initial velocity or position fixes the constants in the corresponding components.

  13. Recall the power rule for differentiation

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

    Each polynomial term is differentiated by reducing the power by one.

  14. Recall the power rule for integration

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

    Each polynomial term is integrated by raising the power by one.

  15. State the velocity vector

    v=6t2−3i+4t3−2j\mathbf{v}=6 t^{2} - 3\mathbf{i}+4 t^{3} - 2\mathbf{j}

    This is the velocity as a function of time.

Answer
v=6t2−3i+4t3−2j\mathbf{v}=6 t^{2} - 3\mathbf{i}+4 t^{3} - 2\mathbf{j}
Question 5
8 markschallenging
A particle moves in a plane with velocity v=6t2−6i+4t−1j\mathbf{v} = 6 t^{2} - 6\mathbf{i}+4 t - 1\mathbf{j} (m s−1^{-1}) at time tt seconds. When t=0t = 0 its position vector is 0i+0j0\mathbf{i} + 0\mathbf{j} m. Find its position vector r\mathbf{r} as a function of tt.
Show worked solution

Worked solution

  1. Write down the velocity vector

    v=6t2−6i+4t−1j\mathbf{v}=6 t^{2} - 6\mathbf{i}+4 t - 1\mathbf{j}

    Position is found by integrating velocity with respect to time.

  2. Integrate the i\mathbf{i} component

    ∫vx dt=2t3−6t+c1\int v_x\,dt=2 t^{3} - 6 t+c_1

    Integrate the i\mathbf{i} component of velocity.

  3. Integrate the j\mathbf{j} component

    ∫vy dt=2t2−t+c2\int v_y\,dt=2 t^{2} - t+c_2

    Integrate the j\mathbf{j} component of velocity.

  4. Use the initial position to find the constants

    c1=0,c2=0c_1=0,\quad c_2=0

    Substitute t=0t = 0 and the given position components.

  5. Recall the vector kinematics definitions

    v=drdt,a=dvdt\mathbf{v}=\frac{d\mathbf{r}}{dt},\quad \mathbf{a}=\frac{d\mathbf{v}}{dt}

    Velocity is the derivative of position and acceleration is the derivative of velocity.

  6. Recall that integration reverses differentiation

    v=∫a dt,r=∫v dt\mathbf{v}=\int \mathbf{a}\,dt,\quad \mathbf{r}=\int \mathbf{v}\,dt

    Integrating acceleration gives velocity, and integrating velocity gives position.

  7. Differentiate each component separately

    ddt(f(t)i+g(t)j)=f′(t)i+g′(t)j\frac{d}{dt}\bigl(f(t)\mathbf{i}+g(t)\mathbf{j}\bigr)=f'(t)\mathbf{i}+g'(t)\mathbf{j}

    With i and j fixed, differentiate the scalar coefficient of each unit vector.

  8. Integrate each component separately

    ∫(f(t)i+g(t)j) dt=(∫f dt)i+(∫g dt)j\int\bigl(f(t)\mathbf{i}+g(t)\mathbf{j}\bigr)\,dt=\Bigl(\int f\,dt\Bigr)\mathbf{i}+\Bigl(\int g\,dt\Bigr)\mathbf{j}

    Each component is integrated independently, each with its own constant.

  9. Recall how speed is defined

    speed=∣v∣=vx2+vy2\text{speed}=\lvert\mathbf{v}\rvert=\sqrt{v_x^{2}+v_y^{2}}

    Speed is the magnitude of the velocity vector, not a component.

  10. Recall how bearing is measured

    bearing is measured clockwise from north\text{bearing is measured clockwise from north}

    In navigation, bearing 000∘000^\circ is due north and 090∘090^\circ is due east.

  11. Use the east and north components for bearing

    bearing=tan⁡−1 ⁣(vxvy) adjusted for quadrant\text{bearing}=\tan^{-1}\!\left(\dfrac{v_x}{v_y}\right)\text{ adjusted for quadrant}

    With i\mathbf{i} east and j\mathbf{j} north, the bearing follows from the velocity components.

  12. Remember constants of integration in each component

    v(t)=(∫ax dt+c1)i+(∫ay dt+c2)j\mathbf{v}(t)=\Bigl(\int a_x\,dt + c_1\Bigr)\mathbf{i}+\Bigl(\int a_y\,dt + c_2\Bigr)\mathbf{j}

    Each component of an indefinite integral needs its own constant, fixed by initial conditions.

  13. Apply initial conditions to each component

    substitute t=t0 into each component separately\text{substitute } t=t_0 \text{ into each component separately}

    Initial velocity or position fixes the constants in the corresponding components.

  14. Recall the power rule for differentiation

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

    Each polynomial term is differentiated by reducing the power by one.

  15. State the position vector

    r=2t3−6ti+2t2−tj\mathbf{r}=2 t^{3} - 6 t\mathbf{i}+2 t^{2} - t\mathbf{j}

    This is the position vector as a function of time.

Answer
r=2t3−6ti+2t2−tj\mathbf{r}=2 t^{3} - 6 t\mathbf{i}+2 t^{2} - t\mathbf{j}

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