It is ok if I post more multiple choice questions in this thread? I'm going through a practice exam at the moment.

**1. The braking distance of a car under constant acceleration in a straight line will be proportional to the square of its initial speed. True or False?**

I used the motion equations (\(\displaystyle v^2 = u^2+2ax\)) but I could not establish a proportion, is there actually one?

**2. The gravitational potential energy of all stationary objects on the surface of the earth is negative. True or False?**

Right now I'm confused about the definition of potential energy. On the one hand I have learnt about it described as mgh. On the other hand, it is supposedly 0 at infinity and negative anywhere nearer.

**3. a) A yacht (Anthea) sees another yacht (Berthilde) sailling at a relative velocity of \(\displaystyle 10ms^{-1}\) directly west. Berthilde is actually sailing NW at the same speed as Anthea. What direction is Anthea sailing in?**

**A. NE**

**B. SE**

**C. SW**

**D. NW**

**E. none of the proceeding**

I think I'm more confused at the way the question is worded than anything. If Anthea believes she is heading north, then would the answer be A?

**3. b) Which of the following is closest to the speed Anthea is sailing at relative to Berthilde?**

**A. 14m/s**

**B. 12m/s**

**C. 10m/s**

**D. 8m/s**

**E. 6m/s**

dunno....

thanks so much!

1. What is the final speed in this case?

2. The gravitational potential energy is given approximately mgh near the Earth's surface. The Newtonian gravity formula gives a better form:

\(\displaystyle U = -\frac{Gm_1m_2}{r}\)

(where r is the distance between the centers of the mass and the Earth.)

However, the point of the question is that we may select any point as the 0 level for the GPE so depending on where you set this level, the GPE could be negative, zero, or positive.

3. Call the velocity of the Berthilde according to the Anthea \(\displaystyle \vec{B}\). Then we know that Anthea is has a velocity \(\displaystyle \vec{v}\) measured in reference to stationary water. Thus the velocity of the Berthilde in reference to the stationary water is given by

\(\displaystyle \vec{v_B} = \vec{B} + \vec{v}\)

Putting the vector diagram together gives a heading vaguely NW.

The other way you can see this is that if the Berthilde is moving due West according to the Anthea, they must have the same Northward component of velocity. And since the Anthea has a Westward component and the Berthilde is moving West relative to it, the Berthilde must also have a Westward component of velocity. Hence the Berthilde is moving NW.

For the last part, we know that the speed of the Berthilde relative to the Anthea is 10 m/s due West. Thus the speed of the Anthea relative to the Berthilde is exactly opposite of this: 10 m/s due East.

-Dan