Explanation:
Displacement is simply the change in position. So in the first part of problem 1, looking at the graph between 0 s and 2 s, the position changes from 0 m to -4 m. So the displacement is:
Δx = -4 m − 0 m
Δx = -4 m
Between 2 s and 4 s, the position stays at -4 m. The displacement is:
Δx = -4 m − (-4 m)
Δx = 0 m
Finally, between 4 s and 6 s, the position goes from -4 m to 6 m. The displacement is:
Δx = 6 m − (-4 m)
Δx = 10 m
The net displacement is the change in position from 0 s to 6 s:
Δx = 6 m − 0 m
Δx = 6 m
In the second part of problem 1, we have a velocity vs time graph.
Car 1 starts with 0 velocity and ends with a velocity of 6 m/s, so it is accelerating and constantly moving to the right.
Car 2 starts with a velocity of -6 m/s and ends with a velocity of 6 m/s. It is also accelerating, but first it is moving to the left, comes to a stop at t = 3 s, then moves to the right.
Car 3 starts with a velocity of 2 m/s and ends with a velocity of 2 m/s. So it is moving constantly to the right, but never speeds up or slows down.
We want to know when two of the cars meet. Unfortunately, this isn't as easy as looking for where the lines cross on the graph. We need to calculate their displacements. We can do this by finding the area under the graph (assuming all the cars start from the same point).
Let's start with Car 2. Half of the area is below the x-axis, and half is above. Without doing calculations, we can say the total displacement for this car is 0. This means it ends back up where it started, and that it never meets either of the other cars, both of which have positive displacements.
So we know Car 1 and Car 3 meet, we just have to find where and when. For Car 1, the area under the curve is a triangle. So its displacement is:
Δx = ½ t v(t)
where t is the time and v(t) is the velocity of Car 1 at that time. Since the line has a slope of 1 and y intercept of 0, we know v(t) = t. So:
Δx = ½ t²
Now look at Car 3. The area under the curve is a rectangle. So its displacement is:
Δx = 2t
When the two cars have the same displacement:
½ t² = 2t
t² = 4t
t² − 4t = 0
t (t − 4) = 0
t = 0, 4
t = 0 refers to the time when both cars are at the starting point, so t = 4 is the answer we're looking for. Where are the cars at this time? Simply plug in t = 4 into either of the equations we found:
Δx = 8
So Cars 1 and 3 meet at 4 s and 8 m.
is percentage of the equity which is to be calculated
is the Risk premium whose value as seen from the attached data for the period 1926-2015 is 8.30%
is the risk aversion factor which is given as 4.
is the standard deviation of the portfolio which from the data for the period 1926-2015 is 20.59
