A-Level Further kinematics Practice Questions

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

velocitydifferentiationaccelerationsubstitutionspeedmagnitude
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A particle moves in a plane so that its position vector at time tt seconds is r=t2i+3tj\mathbf{r} = t^{2}\mathbf{i}+3 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=t2i+3tj\mathbf{r}=t^{2}\mathbf{i}+3 t\mathbf{j}

    The motion is described by this position vector; velocity is its rate of change.

  2. Differentiate each component with respect to time

    drdt=2ti+3j\frac{d\mathbf{r}}{dt}=2 t\mathbf{i}+3\mathbf{j}

    Differentiating each component gives the velocity vector.

  3. State the velocity vector

    v=2ti+3j\mathbf{v}=2 t\mathbf{i}+3\mathbf{j}

    This is the velocity as a function of time.

Answer
v=2ti+3j\mathbf{v}=2 t\mathbf{i}+3\mathbf{j}
Question 2
2 markseasy
A particle has velocity v=6ti+8tj\mathbf{v} = 6t\mathbf{i} + 8t\mathbf{j} m s−1^{-1}. What is its speed at t=1t = 1?
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. Find the speed as the magnitude of velocity

    speed=62+82=10\text{speed}=\sqrt{6^{2}+8^{2}}=10

    At t=1t=1, v=6i+8j\mathbf{v}=6\mathbf{i}+8\mathbf{j}, so speed =36+64=10=\sqrt{36+64}=10.

  3. State the correct choice

    correct choice: 10 m s−1\text{correct choice: } 10\text{ m s}^{-1}

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

Answer
10 m s−110\text{ m s}^{-1}
Question 3
3 marksintermediate
A particle moves with velocity v=3i−3j\mathbf{v} = 3\mathbf{i} - 3\mathbf{j} m s−1^{-1}. Taking i\mathbf{i} as east and j\mathbf{j} as north, what is the bearing of its motion?
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 velocity components to find the bearing

    vx=3, vy=−3⇒bearing=135∘v_x=3,\ v_y=-3 \Rightarrow \text{bearing}=135^\circ

    Eastward and southward motion gives a bearing of 135∘135^\circ (southeast).

  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. State the correct choice

    correct choice: 135∘\text{correct choice: } 135^\circ

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

Answer
135∘135^\circ
Question 4
5 markshard
A particle has position r=(t3−3t)i+(2t2)j\mathbf{r} = (t^{3}-3t)\mathbf{i} + (2t^{2})\mathbf{j} metres. Which is its velocity at t=2t = 2?
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. Differentiate and substitute t=2t = 2

    v=(3t2−3)i+4tj⇒9i+8j\mathbf{v}=(3t^{2}-3)\mathbf{i}+4t\mathbf{j}\Rightarrow 9\mathbf{i}+8\mathbf{j}

    Differentiating gives v=(3t2−3)i+4tj\mathbf{v}=(3t^2-3)\mathbf{i}+4t\mathbf{j}; at t=2t=2 this is 9i+8j9\mathbf{i}+8\mathbf{j}.

  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. State the correct choice

    correct choice: 9i+8j\text{correct choice: } 9\mathbf{i}+8\mathbf{j}

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

Answer
9i+8j9\mathbf{i} + 8\mathbf{j} m s−1^{-1}
Question 5
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}

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