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

2.5.1 SUVAT in 2D

Vectors in Kinematics

The SUVAT equations can be extended to two dimensions. However, strictly speaking, there is no such thing as "squared" vectors, so any equations with squared vector quantities are invalid.

Kinematics Equations:

• Motion in a straight line (1D):
• $v = u + at$
• $s = ut + \frac{1}{2}at^2$
• $s = \frac{1}{2}(u+v)t$
• $v^2 = u^2 + 2as$
• $s = vt - \frac{1}{2}at^2$
• Motion in two dimensions (2D):
• $\mathbf{v} = \mathbf{u} + \mathbf{a}t$
• $\mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2$
• $\mathbf{s} = \frac{1}{2}(\mathbf{u}+\mathbf{v})t$
• $\mathbf{s} = \mathbf{v}t - \frac{1}{2}\mathbf{a}t^2$

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Problem Statement:

Solution:

Given:

• Initial position vector $\mathbf{s_0} = (3, 7)$
• Velocity vector $\mathbf{u} = \mathbf{v} = (2, -1)$ (since velocity is constant)
• Acceleration vector $\mathbf{a} = (0, 0)$
• Time $t = 4$

Part (a):

To find the position vector $\mathbf{s}$:

$$\mathbf{s} = \mathbf{ut} + \frac{1}{2}\mathbf{a}t^2 \Rightarrow \mathbf{s} = \left(2, -1\right)(4) + \frac{1}{2}\left(0, 0\right)(4)^2 = \left(8, -4\right)$$

Remember, we didn't start at ($0, 0$):

$$\text{Position} = \left(3, 7\right) + \left(8, -4\right) = \left(11, 3\right)$$

Part (b):

• Context: Due east means $y = 0$ and $x$ has a value, say $k$.

In part (a), we worked out a position from a known time. We must adapt this method for an unknown time, t.

Equation:

$$\mathbf{r} = \text{Initial Position} + \text{Displacement}$$ $$\mathbf{r} = \left( \begin{array}{c} 3 \\ 7 \end{array} \right) + \left( \begin{array}{c} 2 \\ -1 \end{array} \right)t + \frac{1}{2} \left( \begin{array}{c} 0 \\ 0 \end{array} \right)t^2 = \left( \begin{array}{c} 3 + 2t \\ 7 - t \end{array} \right) \text{ m}$$

• Since $y = 0$, we set:

$$7 - t = 0 \Rightarrow t = :highlight[7 \text{ seconds}]$$

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Additional Problem:

Given:

• $\mathbf{s} = ?$
• $\mathbf{u} = \left( -3, 1 \right)$
• $\mathbf{v} = ?$
• $\mathbf{a} = \left( 2, 3 \right)$
• $t = 3$ Velocity:

$$\mathbf{v} = \mathbf{u} + \mathbf{a}t$$ $$\mathbf{r} = \left( \begin{array}{c} -3 \\ 1 \end{array} \right) + \left( \begin{array}{c} 2 \\ 3 \end{array} \right)^{(3)}= \left( \begin{array}{c} -3 + 6 \\ 1 + 9 \end{array} \right)=\left( \begin{array}{c} 3 \\ 10 \end{array} \right)$$

• Magnitude of velocity:

$$|\mathbf{v}| = \sqrt{3^2 + 10^2} = :highlight[\sqrt{109} \text{ m s\(^{-1}\)}]$$

Bearing:

$$\tan \theta = \frac{10}{3} \Rightarrow \theta = 73.301^\circ$$

• Final Bearing:

$$\therefore \text{Bearing} = :highlight[16.7^\circ]$$

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Problem: Ice Skater on an Ice Rink

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Solution:

Part (a): Acceleration of the Ice Skater

Given:

• Initial velocity $\mathbf{u} = \left( \begin{array}{c} 2.4 \\ -0.6 \end{array} \right)$
• Final velocity $\mathbf{v} = \left( \begin{array}{c} -5.6 \\ 3.4 \end{array} \right)$
• $\mathbf{a}=?$
• Time $t = 20 s$ Using the equation $\mathbf{v} = \mathbf{u} + \mathbf{a}t$:

$$\left( \begin{array}{c} -5.6 \\ 3.4 \end{array} \right) = \left( \begin{array}{c} 2.4 \\ -0.6 \end{array} \right) + \mathbf{a}(20)$$ $$\Rightarrow \left( \begin{array}{c} -8 \\ 4 \end{array} \right) = 20\mathbf{a}$$ $$\Rightarrow \mathbf{a} = :success[\left( \begin{array}{c} -0.4 \\ 0.2 \end{array} \right) \text{ ms\(^{-2}\)}]$$

Part (b): Expression for $\mathbf{s}$ in terms of $t$

Using the equation $\mathbf{s} = \mathbf{ut} + \frac{1}{2}\mathbf{a}t^2$:

$$\mathbf{s} = \left( \begin{array}{c} 2.4 \\ -0.6 \end{array} \right)t + \frac{1}{2}\left( \begin{array}{c} -0.4 \\ 0.2 \end{array} \right)t^2= \left( \begin{array}{c} 2.4t-0.2t^2 \\ -0.6t+0.1t^2 \end{array} \right)$$ $$\Rightarrow \mathbf{s} = :success[\left( \begin{array}{c} 2.4t-0.2t^2 \\ -0.6t+0.1t^2 \end{array} \right) \text{ m}]$$

Part (c): Time at which the skater is directly north-east of $O$

North-east means the x-coordinate equals the y-coordinate, i.e., $2.4t - 0.2t^2 = -0.6t + 0.1t^2$:

$$2.4t - 0.2t^2 = -0.6t + 0.1t^2$$ $$\Rightarrow 0.3t^2 - 3t = 0$$ $$\Rightarrow t(0.3t - 3) = 0 \Rightarrow 0.3t = 3 \Rightarrow t = :highlight[10 \text{ s}]$$

Part (d): Show that the two skaters will meet

For the skaters to meet, their position vectors must be equal at the same time.

Given:

$$\mathbf{r_1} = \left( \begin{array}{c} 2.4t-0.2t^2 \\ -0.6t+0.1t^2 \end{array} \right)$$ $$\mathbf{r_2} = \left( \begin{array}{c} 0 \\ 1.1t-6 \end{array} \right)$$

To determine when the skaters meet, we need to find the time when the x-coordinates are equal:

Time when x-coordinates are equal:

$2.4t - 0.2t^2 = 0$

$\Rightarrow t(2.4 - 0.2t) = 0$

$\Rightarrow 2.4 = 0.2t \Rightarrow t = :highlight[12 \text{ s}]$

Now, check whether this time gives equal y-coordinates:

At $t = 12 s$:

• y-coordinate of skater 1: $y_1 = -0.6(12) + 0.1(12)^2 = 7.2$

• y-coordinate of skater 2: $y_2 = 1.1(12) - 6 = 7.2$

Since both y-coordinates are equal at $t = 12 s$, the skaters meet at the position $\left( \begin{array}{c} 0 \\ 7.2 \end{array} \right)m$ at $t = 12 s.$

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