Variable Acceleration - 2D (AQA A-Level Mathematics): Revision Notes

2.4.1 Using Calculus in 2D

Vectors in Kinematics with Non-Constant Acceleration

When acceleration is non-constant, we cannot use SUVAT equations. We must use calculus.

Relationships:

$$\frac{d\mathbf{s}}{dt} = \mathbf{v} \quad \text{(velocity)} \quad \text{and} \quad \frac{d\mathbf{v}}{dt} = \mathbf{a} \quad \text{(acceleration)}$$ $$\mathbf{s} = \int \mathbf{v} \, dt \quad \text{(displacement)} \quad \text{and} \quad \mathbf{v} = \int \mathbf{a} \, dt$$

Problem 1:

Solution:

(a): Velocity of P

Velocity $\mathbf{v} = \frac{d\mathbf{r}}{dt}$:

$$\mathbf{v} = \frac{d}{dt} \left(3t - 4\right)\mathbf{i} + \frac{d}{dt} \left(t^3 - 4t\right)\mathbf{j} = \left(3\right)\mathbf{i} + \left(3t^2 - 4\right)\mathbf{j} \\ let \ t = 3:\\ \\ \mathbf{v} = 3\mathbf{i} + \left(3(3^2) - 4\right)\mathbf{j} = \left(3\mathbf{i} + 23\mathbf{j}\right) \text{ m s\(^{-1}\)}$$

(b): Acceleration of P

Acceleration $\mathbf{a} = \frac{d\mathbf{v}}{dt}$:

$$\mathbf{a} = \frac{d}{dt} \left(3\right)\mathbf{i} + \frac{d}{dt} \left(3t^2 - 4\right)\mathbf{j} = \left(0\right)\mathbf{i} + \left(6t\right)\mathbf{j}$$

At t = 3:

$$\mathbf{a} = \left(0\mathbf{i} + 18\mathbf{j}\right) \text{ m s\(^{-2}\)}$$

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Problem 2:

Solution:

Velocity $\mathbf{v}$ is given by:

$$\mathbf{v} = \left(t^2 \mathbf{i} + (2t - 3)\mathbf{j}\right)$$

To find acceleration $\mathbf{a}$:

$$\mathbf{a} = \frac{d\mathbf{v}}{dt} = \frac{d}{dt} \left(t^2 \mathbf{i} + (2t - 3)\mathbf{j}\right) = \left(2t\right)\mathbf{i} + \left(2\right)\mathbf{j}$$

At $t = 4$:

$$\mathbf{a} = \left(2(4)\right)\mathbf{i} + \left(2\right)\mathbf{j} = \left(8\mathbf{i} + 2\mathbf{j}\right) \text{ m s\(^{-2}\)}$$

Using $\mathbf{F} = m\mathbf{a}$ and mass $m =$ $3 \times 10^{-3} kg$:

$$\mathbf{F} = 0.003 \times \left(8\mathbf{i} + 2\mathbf{j}\right) = \left(0.024\mathbf{i} + 0.006\mathbf{j}\right) \text{ N}$$

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Problem 3:

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Solution:

Part (a): Distance of $P$ from $O$ when $t = 0$

Given:

• $\mathbf{v} = \left(3t^2 + 2t\right)\mathbf{i} + \left(3t^2 - 4t\right)\mathbf{j}$ To find the position vector $\mathbf{s}, :highlight[integrate] \mathbf{v}$ with respect to $t$:

$$\mathbf{s} = \int \mathbf{v} \, dt = \left(\frac{t^3 + t^2}{3} + \frac{2t^2}{2}\right)\mathbf{i} + \left(\frac{t^3}{3} - 2t^2\right)\mathbf{j} + \mathbf{C}$$

When $t = 2, \mathbf{s} = 9\mathbf{j} m$:

$$\left(0, 9\right) = \left(\frac{2^3 + 2^2}{3} + 2(2^2)\right)\mathbf{i} + \left(\frac{2^3}{3} - 2(2^2)\right)\mathbf{j} + \mathbf{C}$$ $$\mathbf{C} = \left(-12, 5\right)$$

Thus, the position vector $\mathbf{s}$ is:

$$\mathbf{s} = \left(\frac{t^3 + t^2}{3} + 2t^2 - 12\right)\mathbf{i} + \left(\frac{t^3}{3} - 2t^2 + 5\right)\mathbf{j}$$

At $t = 0, \mathbf{s} = \left(-12, 5\right)$.

Distance from $O$:

$$|\mathbf{s}| = \sqrt{(-12)^2 + 5^2} = \sqrt{144 + 25} = :success[13 \text{ m}]$$

Part (b): Acceleration of P at the instant when it is moving parallel to the vector $\mathbf{i}$

Parallel to $\mathbf{i}$ means $\mathbf{v}$ has the form $k(1, 0)$, i.e., the $j-component$ of $\mathbf{v}$ is $0$.

Given $\mathbf{v} = \left(3t^2 + 2t\right)\mathbf{i} + \left(3t^2 - 4t\right)\mathbf{j}$:

Set the $j-component$ to $0$:

$$3t^2 - 4t = 0 \Rightarrow t(3t - 4) = 0 \Rightarrow t = 0 \text{ or } t = \frac{4}{3}$$

To find acceleration, differentiate $\mathbf{v}$ with respect to $t$:

$$\mathbf{a} = \frac{d\mathbf{v}}{dt} = \frac{d}{dt} \left(3t^2 + 2t\right)\mathbf{i} + \frac{d}{dt} \left(3t^2 - 4t\right)\mathbf{j} = \left(6t + 2\right)\mathbf{i} + \left(6t - 4\right)\mathbf{j}$$

Substitute $t = \frac{4}{3}$ to find $\mathbf{a}$ at that instant:

$$\mathbf{a} = \left(6\left(\frac{4}{3}\right) + 2\right)\mathbf{i} + \left(6\left(\frac{4}{3}\right) - 4\right)\mathbf{j} = \left(8 + 2\right)\mathbf{i} + \left(8 - 4\right)\mathbf{j} = :success[\left(10, 4\right) \text{ ms\(^{-2}\)}]$$

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