### Video Transcript

The velocity and height above ground level of an object is shown at different times
in the following figure. The figure is not to scale. The mechanical energy of the object is constant. Which of the following have the same magnitude? Approximate all the velocities to the nearest meter per second. (A) The absolute value of π£ sub three minus the absolute value of π£ sub four and
the absolute value of π£ sub one minus the absolute value of π£ sub two. (B) The absolute value of π£ sub three minus the absolute value of π£ sub one and 10
meters per second minus the absolute value of π£ sub three. (C) The absolute value of π£ sub two minus the absolute value of π£ sub four and the
absolute value of π£ sub three minus the absolute value of π£ sub one. (D) The absolute value of π£ sub three minus the absolute value of π£ sub one and the
absolute value of π£ sub one minus the absolute value of π£ sub two.

In this question, we are given a figure showing the motion of an object as it moves
along a path of various heights. We want to calculate the velocity of the object at each interval its height was
measured at so that we can determine which of the answer options have the same
magnitude. Before we begin to solve this, we need to recall some information about mechanical
energy and how we can use it to calculate the values that we want.

We can recall that the mechanical energy, ME, of an object is defined as the sum of
its kinetic energy and potential energy. The kinetic energy, KE, is equal to one-half multiplied by the mass, π, multiplied
by the velocity, π£, squared. We also have gravitational potential energy, GPE, in this problem, which is equal to
the mass, π, multiplied by the acceleration due to gravity, π, multiplied by the
height of the object, β. Therefore, the mechanical energy of the object is equal to one-half ππ£ squared plus
ππβ.

We want to rearrange this equation to make the velocity π£ the subject. We will clear some space for working out. But before we do, we will note down that the question tells us that the mechanical
energy of the object is constant in the figure. Now that we have noted this down, letβs clear some space.

We start by subtracting ππβ from both sides to give us ME minus ππβ equals
one-half ππ£ squared. We then divide both sides by π over two, giving us two multiplied by ME over π
minus πβ equals π£ squared. We then take the square root of both sides to leave us with π£ equals the square root
of two multiplied by ME over π minus πβ. Before we calculate the individual velocities, we can calculate a value for ME over
π. We are told in the question that the mechanical energy of the object is constant, but
another constant in this system is the mass of the object. So if we divide both sides of the mechanical energy equation by the mass, π, we find
that ME over π is equal to one-half π£ squared plus πβ.

We can see from the figure that the initial velocity of the object is 10 meters per
second and the initial height is 25 meters. We can also recall that the acceleration due to gravity, π, is equal to 9.8 meters
per second squared. Substituting these values into this equation and grabbing a calculator, we find that
ME over π is equal to 295 meters squared per second squared. Remember that because the mechanical energy and mass are constants, the quantity ME
over π is going to be the same at every position.

We can now go ahead and calculate the individual velocities using the equation we
found above. From the figure, we can see that the height at position one is 15 meters. So substituting this into our velocity equation and grabbing a calculator, we find
that π£ sub one is equal to 17 meters per second to the nearest meter per
second. We can see that the height at position two is 10 meters, so π£ sub two will be equal
to 20 meters per second. Similarly, we can see that the height at position three is 20 meters, so π£ sub three
will be equal to 14 meters per second. And the height at position four is zero meters, so π£ sub four will be equal to 24
meters per second.

Now that we know the magnitudes of all our velocities, letβs look at the answer
options we have been given and calculate the values that we need.

Letβs begin with option (A). The absolute value of π£ sub three minus the absolute value of π£ sub four is equal
to 14 meters per second minus 24 meters per second, which equals negative 10 meters
per second. The absolute value of π£ sub one minus the absolute value of π£ sub two is equal to
17 meters per second minus 20 meters per second, which equals negative three meters
per second. We can see that these two expressions do not have the same magnitude, so option (A)
is incorrect.

Now letβs consider option (B). The absolute value of π£ sub three minus the absolute value of π£ sub one is equal to
14 meters per second minus 17 meters per second, which equals negative three meters
per second. 10 meters per second minus the absolute value of π£ sub three is equal to 10 meters
per second minus 14 meters per second, which equals negative four meters per
second. We can see that these two expressions do not have the same magnitude, so option (B)
is incorrect.

Now letβs look at option (C). The absolute value of π£ sub two minus the absolute value of π£ sub four is equal to
20 meters per second minus 24 meters per second, which equals negative four meters
per second. We can see that this is not equal to the absolute value of π£ sub three minus the
absolute value of π£ sub one, so option (C) is incorrect.

This leaves us with option (D). If we now compare the absolute value of π£ sub three minus the absolute value of π£
sub one with the absolute value of π£ sub one minus the absolute value of π£ sub
two, we can see that both of these expressions are equal to negative three meters
per second. These have the same magnitude. So therefore, option (D) must be the correct answer. The absolute value of π£ sub three minus the absolute value of π£ sub one is equal to
the absolute value of π£ sub one minus the absolute value of π£ sub two.