Motion using Vectors HELP (1 Viewer)

csi · Mar 9, 2021

A small finishing boat can travel at 8km/h in still water. It is being steered due east, but there is a current running south at 2km/h and a breeze blowing the boat south west at 4km/h. Find the resultant velocity of the boat, giving the speed correct to 2dp and the direction correct to nearest degree.
ANS: 7.08 km/h, 133 degrees

If possible, please show me how to draw the diagram representing the above information. This would help me to understand the question a lot better. THANKS!

csi · Mar 9, 2021

HaydenChau said:
View attachment 30366 View attachment 30367

tysm

Directrix · Mar 9, 2021

HaydenChau said:
View attachment 30366 View attachment 30367

Can I book a weekly checkup with you, Dr Chau?

CM_Tutor · Mar 9, 2021

Using Vectors

Define the directions of the $\overrightarrow{i}$ and $\overrightarrow{j}$ vectors as East and North, respectively.

Consequently, the vector for the velocity of the boat heading due East (in still water) is $8\overrightarrow{i}$ .

The vector for the velocity relating to the current is $-2\overrightarrow{j}$ .

The velocity relating to the breeze is 4 km/h in the direction South-West. This direction is $-\overrightarrow{i} - \overrightarrow{j}$ , and the required vector is thus $-k\left(\overrightarrow{i} + \overrightarrow{j}\right)$ , where $k>0$ , and must have a magnitude of 4. That is,

$\begin{align*} \left(-k\right)^2 + \left(-k\right)^2 &= 4^2 \\ 2k^2 &= 16 \\ k^2 &= 8 \\ k &= 2\sqrt{2} \qquad \text{as $k>0$} \end{align*}$

So, the vector for the velocity relating to the current is $-2\sqrt{2}\left(\overrightarrow{i} + \overrightarrow{j}\right)$ .

It follows that the net velocity vector for the boat is

$\overrightarrow{v} = 8\overrightarrow{i} - 2\overrightarrow{j} - 2\sqrt{2}\left(\overrightarrow{i} + \overrightarrow{j}\right) = 2\left(4 - \sqrt{2}\right)\overrightarrow{i} - 2\left(1 + \sqrt{2}\right)\overrightarrow{j}$ .

The speed of the boat is thus:

$\begin{align*} \left|\overrightarrow{v}\right| &= \left|2\left(4 - \sqrt{2}\right)\overrightarrow{i} - 2\left(1 + \sqrt{2}\right)\overrightarrow{j}\right| \\ \left|\overrightarrow{v}\right|^2 &= \left[2\left(4 - \sqrt{2}\right)\right]^2 + \left[-2\left(1 + \sqrt{2}\right)\right]^2 \\ &= 4\left(4^2 - 8\sqrt{2} + 2\right) + 4\left(1 + 2\sqrt{2} + 2\right) \\ &= 4\times 18 -32\sqrt{2} + 4\times 3 + 8\sqrt{2} \\ &= 84 - 24\sqrt{2} \\ &= 4\left(21 - 6\sqrt{2}\right) \\ \left|\overrightarrow{v}\right| &= 2\sqrt{21-6\sqrt{2}} \\ &\approx 7.0752... \text{ km/h} \end{align*}$

The direction is $\text{S}\theta^o\text{E}$ where

$\tan\theta = \frac{2\left(4-\sqrt{2}\right)}{2\left(1+\sqrt{2}\right)} = \frac{\left(4-\sqrt{2}\right)\left(1-\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)} = \frac{6-5\sqrt{2}}{1-2} = 5\sqrt{2}-6$

$\theta = \tan^{-1}\left(5\sqrt{2}-6\right) \approx 46.965...^o$

So, the speed is 7.08 km/h in direction $\text{S}47^o\text{E}$ (or $133^o\text{T}$ ).

Directrix · Mar 9, 2021

HaydenChau said:
My writing hahahaha?

No it's pretty decent tbh... probably a gp level not a full on doctor

Imagine having the worst handwriting and being too dumb to do med... me 1000000%

csi · Mar 9, 2021

CM_Tutor said:
Using Vectors

Define the directions of the $\overrightarrow{i}$ and $\overrightarrow{j}$ vectors as East and North, respectively.

Consequently, the vector for the velocity of the boat heading due East (in still water) is $8\overrightarrow{i}$ .

The vector for the velocity relating to the current is $-2\overrightarrow{j}$ .

The velocity relating to the breeze is 4 km/h in the direction South-West. This direction is $-\overrightarrow{i} - \overrightarrow{j}$ , and the required vector is thus $-k\left(\overrightarrow{i} + \overrightarrow{j}\right)$ , where $k>0$ , and must have a magnitude of 4. That is,

$\begin{align*} \left(-k\right)^2 + \left(-k\right)^2 &= 4^2 \\ 2k^2 &= 16 \\ k^2 &= 8 \\ k &= 2\sqrt{2} \qquad \text{as $k>0$} \end{align*}$

So, the vector for the velocity relating to the current is $-2\sqrt{2}\left(\overrightarrow{i} + \overrightarrow{j}\right)$ .

It follows that the net velocity vector for the boat is $\overrightarrow{v} = 8\overrightarrow{i} - 2\overrightarrow{j} - 2\sqrt{2}\left(\overrightarrow{i} + \overrightarrow{j}\right) = 2\left(4 - \sqrt{2}\right)\overrightarrow{i} - 2\left(1 + \sqrt{2}\right)\overrightarrow{j}$ .

The speed of the boat is thus:

$\begin{align*} \left|\overrightarrow{v}\right| &= \left|2\left(4 - \sqrt{2}\right)\overrightarrow{i} - 2\left(1 + \sqrt{2}\right)\overrightarrow{j}\right| \\ \left|\overrightarrow{v}\right|^2 &= \left[2\left(4 - \sqrt{2}\right)\right]^2 + \left[-2\left(1 + \sqrt{2}\right)\right]^2 \\ &= 4\left(4^2 - 8\sqrt{2} + 2\right) + 4\left(1 + 2\sqrt{2} + 2\right) \\ &= 4\times 18 -32\sqrt{2} + 4\times 3 + 8\sqrt{2} \\ &= 84 - 24\sqrt{2} \\ &= 4\left(21 - 6\sqrt{2}\right) \\ \left|\overrightarrow{v}\right| &= 2\sqrt{21-6\sqrt{2}} \\ &\approx 7.0752... \text{ km/h} \end{align*}$

The direction is $\text{S}\theta^o\text{E}$ where

$\tan\theta = \frac{2\left(4-\sqrt{2}\right)}{2\left(1+\sqrt{2}\right)} = \frac{\left(4-\sqrt{2}\right)\left(1-\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)} = \frac{6-5\sqrt{2}}{1-2} = 5\sqrt{2}-6$

$\theta = \tan^{-1}\left(5\sqrt{2}-6\right) \approx 46.965...^o$

So, the speed is 7.08 km/h in direction $\text{S}47^o\text{E}$ (or $133^o\text{T}$ ).

TY!!

Motion using Vectors HELP (1 Viewer)

csi

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