Video Transcript
An aircraft is flying eastward,
accelerating at 12 meters per second squared. The aircraft is also descending
vertically with an acceleration of nine meters per second squared. Before accelerating, the eastward
velocity of the aircraft was 250 meters per second and its vertical speed was 10
meters per second. What is the speed of the aircraft
after 10 seconds of acceleration? Answer to the nearest meter per
second.
Okay, so this question is asking us
about the speed and acceleration of an aircraft. Let’s begin by drawing a diagram of
the situation using the information that we’ve been given. Let’s suppose that this is our
aircraft, and we’re told that it’s flying eastward, which we’ve drawn to the right
in our diagram. We’re told that the aircraft starts
out with an eastward velocity of 250 meters per second. We’ve labeled this as 𝑢 subscript
h, where the h represents that this is the horizontal component of the aircraft’s
velocity.
We’re also told that the aircraft
is descending vertically with an initial vertical speed of 10 meters per second. This is the vertical component of
the aircraft’s initial velocity, and we’ve labeled it as 𝑢 subscript v, where the v
stands for vertical. We can notice that this is directed
vertically downward and has a positive value. This means that we’ve implicitly
taken downward to be the positive vertical direction. Similarly, for the horizontal
motion, we’ve implicitly taken right or eastward to be positive.
The other information that we’re
given is about the aircraft’s acceleration. We’re told that it has an eastward
acceleration, which we’ve labeled as 𝑎 subscript h, of 12 meters per second squared
and that its vertical acceleration as it descends is nine meters per second
squared. We’ve labeled this as 𝑎 subscript
v, and this is also directed vertically downward.
We’re being asked to consider the
speed of the aircraft after it’s been accelerating for 10 seconds. Let’s label this time interval as
Δ𝑡. And we’ll suppose that after
accelerating for this time of Δ𝑡, the aircraft ends up with a horizontal velocity
component of 𝑉 subscript h and a vertical velocity component of 𝑉 subscript v. In order to work out the values of
𝑉 subscript h and 𝑉 subscript v, we can recall that acceleration is defined as the
rate of change of an object’s velocity. So, if an object’s velocity changes
by an amount Δ𝑉 over a time interval of Δ𝑡, then its acceleration 𝑎 is equal to
Δ𝑉 divided by Δ𝑡.
We can apply this equation to the
aircraft’s horizontal motion and its vertical motion separately. For the horizontal motion, the
acceleration is the horizontal acceleration 𝑎 subscript h. And the change in velocity Δ𝑉 is
the final horizontal component of the velocity 𝑉 subscript h minus the initial
component 𝑢 subscript h. The equation for the vertical
motion looks just the same but with the vertical acceleration 𝑎 subscript v and the
vertical velocity components 𝑉 subscript v and 𝑢 subscript v.
To use these two equations to
calculate the final components of the aircraft’s velocity, we’re going to need to
rearrange them in order to make these final velocity terms the subject. Since both equations have exactly
the same form, then the process will be the same in each case. Let’s see how this works using the
horizontal equation. To make 𝑉 subscript h the subject
of this equation, we begin by multiplying both sides by the time interval Δ𝑡. On the right-hand side of the
equation, there’s then a Δ𝑡 in the numerator, which cancels with the Δ𝑡 in the
denominator.
If we write the equation the other
way around, we then have that 𝑉 subscript h minus 𝑢 subscript h is equal to 𝑎
subscript h multiplied by Δ𝑡. From here, we then add the initial
horizontal velocity 𝑢 subscript h to each side. On the left-hand side, the 𝑢
subscript h cancels out with the minus 𝑢 subscript h. So we end up with an equation for
the final horizontal velocity 𝑉 subscript h, which says that it’s equal to the
horizontal acceleration 𝑎 subscript h multiplied by Δ𝑡, the time interval, plus
the initial horizontal velocity 𝑢 subscript h.
And we also know that we’re going
to get exactly the same equation for the vertical motion. But in this case, we’ll have the
vertical velocity components and the vertical acceleration. We’re now in a position to
substitute in our values to the right-hand sides of both of these two equations. Let’s clear ourselves some space so
that we can do this.
Okay, let’s begin with the
horizontal equation. Substituting in our values for the
horizontal acceleration, the time interval, and the initial horizontal velocity, we
get that the final horizontal velocity 𝑉 subscript h is equal to 12 meters per
second squared multiplied by 10 seconds plus 250 meters per second. So that’s the horizontal
acceleration 𝑎 subscript h multiplied by the time interval Δ𝑡 plus the initial
horizontal velocity 𝑢 subscript h.
Let’s now evaluate this expression
for 𝑉 subscript h. The first term is 12 meters per
second squared multiplied by 10 seconds. This works out as 120 meters per
second. And then we have to add the second
term, the 250 meters per second. Adding together these two terms, we
find that the final horizontal velocity is 370 meters per second.
Let’s now do the same thing for the
vertical motion, substituting in our values to this equation for the final vertical
velocity. We find that 𝑉 subscript v is
equal to nine meters per second squared multiplied by 10 seconds plus 10 meters per
second. So that’s the value for the
vertical acceleration 𝑎 subscript v multiplied by the time interval Δ𝑡 plus the
initial vertical velocity 𝑢 subscript v.
This first term on the right-hand
side works out as 90 meters per second. And so we have that the final
vertical velocity 𝑉 subscript v is equal to 90 meters per second plus 10 meters per
second, which is 100 meters per second.
So we now know the horizontal
component and the vertical component of the aircraft’s final velocity. What we want to know is the final
speed of the aircraft. This final speed will be the
magnitude of the aircraft’s resultant final velocity. In order to find the resultant of
two vectors, we need to add those vectors tip to tail. What this means is that in order to
find the resultant final velocity of the aircraft, we can draw the vector for the
horizontal velocity like this. And then we draw in the vector for
the vertical velocity, starting with the tail of this vector at the tip of the
horizontal velocity vector.
The resultant of these two vectors
is then the vector which starts at the tail of the first vector, that’s the
horizontal velocity vector, and extends to the tip of the second vector, the
vertical velocity vector. So this diagonal arrow here is the
vector which represents the final resultant velocity of the aircraft.
Let’s label the magnitude of this
vector, which gives us the final speed of the aircraft, as 𝑉. In order to find the value of 𝑉,
we can make use of Pythagoras’s theorem. This says that, for a general
right-angled triangle with a hypotenuse of length 𝑐 and other sides of lengths 𝑎
and 𝑏, that 𝑐 squared is equal to 𝑎 squared plus 𝑏 squared. Or by taking the square root of
both sides of the equation, we have that the length of the hypotenuse 𝑐 is equal to
the square root of 𝑎 squared plus 𝑏 squared.
In our case, since we’ve got one
vector representing a horizontal velocity component and another representing a
vertical velocity component, then the angle between these two vectors must be 90
degrees so that we have a right-angled triangle. The length of the hypotenuse of
this triangle is the final speed of the aircraft 𝑉, which is what we’re trying to
find. And the other two sides of this
triangle are values that we’ve already calculated for the horizontal and vertical
components of the aircraft’s final velocity.
By applying this equation from
Pythagoras’s theorem to this triangle for the aircraft’s final velocity, we get that
𝑉 is equal to the square root of 𝑉 subscript h squared plus 𝑉 subscript v
squared. Then, substituting in our values
for 𝑉 subscript h and 𝑉 subscript v, we get that 𝑉 is equal to the square root of
the square of 370 meters per second plus the square of 100 meters per second. Taking the square of each of these
two terms and then adding them together, we find that 𝑉 is equal to the square root
of 146900 meters squared per second squared. When we then evaluate the square
root, we get a result for 𝑉 of 383.28 meters per second. The ellipsis here indicates that
this result has further decimal places.
Looking back at the question, we
see that we’re asked to give our answer to the nearest meter per second. By rounding our result, we get our
final answer to the question that after 10 seconds of acceleration, the speed of the
aircraft is 383 meters per second.
