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decent · Jul 12, 2009, 12:00 AM

1. Jack and Rose are out fishing on the lake on a hot summer day when they both decided to go for a swim. Rose dives off from the front of the boat with a force of 55 N, while Jack dives off the boat with a force of 75 N. Find the resultant force on the boat.

2. Ria sees a bird outside her bedroom window. She says the bird moved 40 m East, 30 m North and 60 m West before it disappeared. What was the displacement of the bird while Ria was watching it?

3. Minnie dives off the 4-m springboard and initially bounces up with a velocity of 8 m/s at an angle of 60o to the horizontal. What are the horizontal and vertical components of her velocity?

4. A boat leaves shore and travels 30 km East and then 15 km 50o South of East. Find the boat's resultant displacement vector.

5. Sophia leaves her house. Drives 20 km due North, then turns unto a narrow street and continues in the direction 45o North East for 30 km, and finally turns into the main highway due East for 50 km. What is her total displacement from the house?
Is it 94 km?

Unknown008 · Jul 12, 2009, 12:31 AM

1. For the first question, doesn't it say where does Jack dive from? If he dived from the front, the forces would add up... if he dived from the rear, the forces would subtract... and from a side, you'll have to find the resultant by Pythagoras' theorem.

2. Have a drawing with those types of numbers. Physics is mostly about everyday life things, and sketches or mental pictures would definitely help you. The displacement is the distance between the starting point to the final point, in a straight line. (ans = 36.06 m, bearing of 326 degrees)

3. Have a drawing here too. Here only the speed and the angle to the horizontal are needed (the 4-m springboard is insignificant). In your sketch, you'll have a line, 60 degrees to the horizontal, indicating 8 m/s. The vertical component will be given through trigonometry;

$\frac{The\, horizontal\, component}{hypotenuse} = cos (of\, angle\, between\, it,\, 60)$

$\frac{The\, vertical\, component}{hypotenuse} = sin (of\, angle\, between\, adjacent\, and \, hypotenuse,\, 60)$

(ans: vertical = 6.93 m/s, horizontal = 4 m/s)

4. Could you try this one? It's the same principle as the question no.2 You'll need your cosine rule here;

$a^2 = b^2 + c^2 - 2(b)(c)cosA.$

And your sine rule to find the direction;

$\frac{sin b}{B} = \frac{sin a}{A}$

(ans: 41.3 km, 16.2 degrees South of East or bearing of 106.2 degrees )

Unknown008 · Jul 12, 2009, 02:02 AM

3. Can you now find the vertical and horizontal components?

EDIT: OOps, the 'vertical' didn't show up...