Question 2.
1 answer:
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a) Not possible
b) Yes, it's possible (see graph in attachment)
Explanation:
a)
The average velocity of a body is defined as the ratio between the displacement and the time elapsed:
where
is the displacement
is the time elapsed
In this problem, we want to have zero displacement and non-zero average velocity. From the equation above, we see that this is not possible. In fact, if the total displacement is zero,
And therefore as a consequence,
which means that the average velocity is zero.
B)
Here we want to have zero displacement and non-zero velocity. In this case, it is possible: in fact, we are not talking about average velocity, but we are talking about (instantaneous) velocity.
On a position-time graph, the instantaneous velocity is the slope of the graph. Look at the graph in attachment. We see that the position of the object first increases towards positive value, then it decreases (the object starts moving backward), then becomes negative, then it increases again until returning to the original position, x = 0.
In all of this, we notice that the total displacement of the object is zero:
However, we notice that the instantaneous velocity of the object at the various instants is not zero, because the slope of the graph is not zero.
Learn more about average velocity:
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Answer:
2560m or 2.56km (rounded to 3 significant figures)
Explanation:
First, list all known and desired values/variables (initial vertical velocity is 0 as the plane is kept level and vertical acceleration is just gravity):
The horizontal displacement is going to be the distance travelled, horizontally of course, once the package is released;
First thing to understand is that the vertical and horizontal components are to be dealt with separately because they don't affect each other;
Since there is no horizontal acceleration (ignoring air resistance), we simply require a velocity and time to find the horizontal displacement, using the formula v = d/t (or speed = distance/time);
What we have is the horizontal velocity but we don't have the time taken;
One thing we know is that the time elapsed for the vertical fall of 800m and for the horizontal displacement must be the same;
What we do, therefore, is find the time taken for the vertical displacement using the formula, s = ut + ¹/₂·at², since we know the vertical velocity, height and acceleration:
800 = (0)t + ¹/₂·(9.8)t²
800 = 4.9t²
t² = 163.26...
t = 12.77...
We now have the time taken for the vertical fall and the horizontal displacement, we can use this with the horizontal velocity we know already and get the horizontal displacement: