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alukav5142 [94]

3 years ago

10

After 3 hours, Ernie had travelled 195 miles. If he travels at a constant speed, how far will he have

1 answer:

Umnica [9.8K]3 years ago

7 0

Answer:

x = 325 miles

Step-by-step explanation:

$\frac{195}{3} = \frac{x}{5} \\ 3.x = 5.195 \\ 3x = 975 \\ x = \frac{975}{3} \\ x = 325 \: miles$

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LA plane is trying to travel 250 miles at a bearing of 20° E of S, however, it ends 230 miles away from the

Semmy [17]

Answer:

The wind pushed the plane $65.01$ miles in the direction of $45.56 ^{\circ}$ East of North with respect to the destination point.

Step-by-step explanation:

Let origin, O, br the starting point and point D be the destination at 250 miles at a bearing of 20° E of S, but due to wind let D' be the actual position of the plane at 230 miles away from the starting point in the direction of 35° E of South as shown in the figure.

So, we have |OD|=250 miles and |OD'|=230 miles.

Vector $\overrightarrow{DD'}$ is the displacement vector of the plane pushed by the wind.

From figure, the magnitude of the required displacement vector is

$|DD'|=\sqrt{|AB|^2+|PQ|^2}\;\cdots(i)$

and the direction is $\alpha$ east of north as shown in the figure,

$\tan \alpha=\frac{|PQ|}{|AB|}\;\cdots(ii)$

From the figure,

$|AB|=|OA-OB|$

$\Rightarrow |AB|=|OD\cos 20 ^{\circ}-OD'\cos 35 ^{\circ}|$

$\Rightarrow |AB|=|250\cos 20 ^{\circ}-230\cos 35 ^{\circ}|$

$\Rightarrow |AB|=45.52$ miles

Again, $|PQ|=|OP-OQ|$

$\Rightarrow |PQ|=|OD\sin 20 ^{\circ}-OD'\sin 35 ^{\circ}|$

$\Rightarrow |PQ|=|250\sin 20 ^{\circ}-230\sin 35 ^{\circ}|$

$\Rightarrow |PQ|=46.42$ miles

Now, from equations (i) and (ii), we have

$|DD'|=\sqrt{|45.52|^2+|46.42|^2}=65.01$ miles, and

$\tan \alpha=\frac{|46.42|}{|45.52|}$

$\alpha=\tan^{-1}\left(\frac{|46.42|}{|45.52|}\right)=45.56 ^{\circ}$

Hence, the wind pushed the plane $65.01$ miles in the direction of $45.56 ^{\circ}$ E astof North with respect to the destination point.

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