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3 years ago

6

An object with a starting velocity of 15 m/s accelerates at 3 m/s2. How far does the object travel within 10 seconds?

1 answer:

Anna007 [38]3 years ago

6 0

Answer: 300 m

<u>Step-by-step explanation:</u>

$distance(s)=v_it+\dfrac{1}{2}at^2 \quad \text{where}\ v_i \text{is initial velocity, a is acceleration, t is time}$

$s=(15)(10)+\dfrac{1}{2}(3)(10)^2$

= 150 + 150

= 300

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The bearing of a lighthouse from a ship is N 37° E. The ship sails 2.5 miles further from the lighthouse. The new bearing is 25°

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Answer:

The distance between the ship at N 25°E and the lighthouse would be 7.26 miles.

Step-by-step explanation:

The question is incomplete. The complete question should be

The bearing of a lighthouse from a ship is N 37° E. The ship sails 2.5 miles further towards the south. The new bearing is N 25°E. What is the distance between the lighthouse and the ship at the new location?

Given the initial bearing of a lighthouse from the ship is N 37° E. So, $\angle ABN$ is 37°. We can see from the diagram that $\angle ABC$ would be $180-37=$ 143°.

Also, the new bearing is N 25°E. So, $\angle BCA$ would be 25°.

Now we can find $\angle BAC$ . As the sum of the internal angle of a triangle is 180°.

$\angle ABC+\angle BCA+\angle BAC=180\\143+25+\angle BAC=180\\\angle BAC=180-143-25\\\angle BAC=12$

Also, it was given that ship sails 2.5 miles from N 37° E to N 25°E. We can see from the diagram that this distance would be our BC.

And let us assume the distance between the lighthouse and the ship at N 25°E is $AC=x$

We can apply the sine rule now.

$\frac{x}{sin(143)}=\frac{2.5}{sin(12)}\\ \\x=\frac{2.5}{sin(12)}\times sin(143)\\\\x=\frac{2.5}{0.207}\times 0.601\\ \\x=7.26\ miles$

So, the distance between the ship at N 25°E and the lighthouse is 7.26 miles.

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Step-by-step explanation:

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