Video Transcript
The change in an object’s velocity
with time is shown in the graph. The object has a velocity 𝑢 at the
time 𝑡 one and a velocity 𝑣 at the time 𝑡 two. What quantity is equal to the
quantity velocity multiplied by the quantity time?
In this question, we’re given a
graph that shows the velocity of an object on the 𝑦- or vertical axis against time
on the horizontal or 𝑥-axis. We can see from this graph that at
a time of 𝑡 one, the velocity of the object has a value equal to 𝑢, while at a
time of 𝑡 two, the velocity of the object is equal to 𝑣. This matches the information that
we’re told in the text of the question. In this first part of the question,
we’re asked what quantity is equal to the quantity velocity multiplied by the
quantity time. In order to answer this, we can
recall that the velocity of an object is defined as the rate of change of that
object’s displacement with time. So, an object’s velocity is equal
to the displacement of that object divided by the time over which that displacement
took place.
If we multiply both sides of this
equation by the quantity time, then on the right-hand side, the time in the
numerator cancels with the time in the denominator. So, we find then that time
multiplied by velocity is equal to displacement. So, our answer to the first part of
the question is that the quantity that’s equal to the quantity velocity multiplied
by the quantity time is displacement. We can notice that because
displacement is equal to velocity multiplied by time, then if we have a graph of
velocity against time, like the one in this question, then for the object whose
motion is shown on the graph, the displacement of the object between two times is
equal to the area under the velocity–time graph between those two time values.
Now, let’s clear some space and
move on to the second part of the question.
Which of the following is the
correct expression for the value of the area 𝐵 shown in the graph? (A) 𝑢 divided by Δ𝑡, (B) 𝑣 minus
𝑢 multiplied by Δ𝑡, (C) 𝑢 multiplied by Δ𝑡, (D) Δ𝑡 divided by 𝑢.
We are being asked here to find the
value of the area labeled as 𝐵 on this velocity–time graph. We can see that area 𝐵 is this
green rectangle here on the graph. We can also see that the bottom
left-hand corner of this green rectangle is located at the origin of the graph. So, that’s at a time value of 𝑡
one and a velocity value of zero. On the horizontal time axis, we can
notice that the rectangle extends out from a time of 𝑡 one to a time of 𝑡 two, and
its length along this axis is labeled as Δ𝑡, which is 𝑡 two minus 𝑡 one. If we refer to this dimension as
the width of the rectangle, then we can say that this rectangle has a width equal to
Δ𝑡.
If we now consider the vertical
velocity axis, we can notice that a velocity of 𝑢 is marked at this height at the
top of the rectangle. So, this green rectangle
representing the area 𝐵 extends from a velocity of zero to a velocity of 𝑢. That means then that the height of
the rectangle on the velocity axis is equal to 𝑢 minus zero, which is simply
𝑢. So then, this green rectangle has a
height that’s equal to 𝑢. Let’s recall that the area of a
rectangle is equal to the rectangle’s height multiplied by its width. Area 𝐵 is the area of the green
rectangle with a height of 𝑢 and a width of Δ𝑡. So then, area 𝐵 is equal to 𝑢
multiplied by Δ𝑡.
We can see that this expression we
found for the area 𝐵 matches the expression given to us in option (C). That means that option (C) is our
answer. The correct expression for the
value of the area 𝐵 shown in the graph is 𝑢 multiplied by Δ𝑡.
Let’s once again clear some space
so we can look at the third part of the question.
Which of the following is the
correct expression for the value of the area 𝐴 shown in the graph? (A) 𝑢 plus 𝑣 times Δ𝑡, (B) two
times 𝑢 plus 𝑣 times Δ𝑡, (C) a half times 𝑢 plus 𝑣 times Δ𝑡, (D) a half times
𝑣 minus 𝑢 times Δ𝑡, (E) a half times 𝑢 minus 𝑣 times Δ𝑡.
Okay, so this time, we’re trying to
find the area shown in the graph that’s labeled as area 𝐴. We can see that this is the area of
this yellow triangle here, which is located above the green rectangle that was area
𝐵. Notice that the width of the base
of this triangle or its extent along the time axis is the same as the width of the
green rectangle. Just like the rectangle, it extends
from a time of 𝑡 one out to a time of 𝑡 two. And that gives it a width of 𝑡 two
minus 𝑡 one, which is equal to Δ𝑡. So, we can say then that the length
of the base of this triangle is equal to Δ𝑡.
Let’s now consider the height of
the triangle, so that’s its extent along the vertical velocity axis. We can see that the base of the
triangle is at a height of 𝑢 on this velocity axis. So, the triangle starts at this
height of 𝑢, and we can see that the top corner of the triangle is at a height or
velocity of 𝑣. That means that the triangle has a
height equal to 𝑣 minus 𝑢. And just to be clear, 𝑣 is this
vertical distance from the time axis up to the horizontal line labeled as 𝑣, while
𝑢 is the vertical distance up from the same time axis to the horizontal line that’s
labeled as 𝑢.
So, this length on the vertical
axis that we’ve labeled as 𝑣 minus 𝑢 is equal to this length of 𝑣, which is the
total height on the velocity axis, minus this length of 𝑢, which is the height of
the rectangle which the triangle is on top of. So then, we have that the height of
this triangle is equal to 𝑣 minus 𝑢.
We can recall that the area of a
triangle is equal to a half multiplied by its base multiplied by its height. In this case, it’s straightforward
to see why this factor of a half comes in. We can see that the yellow triangle
on the graph takes up half of the area we’ve marked out by this blue rectangle. We see that we can divide this blue
rectangle into two equal triangles, one of which is the yellow triangle representing
area 𝐴 and the other is this triangle that we’re marking out in green up here.
We can notice that the blue
rectangle has a height of 𝑣 minus 𝑢 and a width of Δ𝑡. We know that the area of a
rectangle is equal to its width multiplied by its height. In this case for the blue
rectangle, that’s Δ𝑡 multiplied by 𝑣 minus 𝑢. Since each of the two triangles in
this rectangle, so that’s the yellow one representing area 𝐴 and the green one that
we marked out, is equal to half of the area of the rectangle, then the area of one
of the triangles must be equal to this width times height for the area of the
rectangle multiplied by a half. This then leads us to this formula
for the area of the triangle as equal to a half multiplied by its base, which is the
same as the width of the rectangle multiplied by its height.
We know that for this yellow
triangle representing area 𝐴, the length of its base is Δ𝑡 and its height is 𝑣
minus 𝑢. Substituting these values into our
formula for the area of a triangle, we get that area 𝐴 is equal to a half
multiplied by Δ𝑡 multiplied by 𝑣 minus 𝑢. When we’re just multiplying
together different terms, the order that we write them in doesn’t matter. So, we can also write this as a
half times 𝑣 minus 𝑢 times Δ𝑡. We can then see more clearly that
this expression matches the one that’s given in option (D). The correct expression for the
value of the area 𝐴 shown in the graph is a half times 𝑣 minus 𝑢 times Δ𝑡.
Let’s now clear some space one last
time and have a look at the final part of the question.
Which of the following formulas
correctly expresses the relationship between acceleration, displacement, and change
in time shown by the graph? (A) 𝑠 equals 𝑢 times Δ𝑡 plus
half times 𝑎 times Δ𝑡 squared. (B) 𝑠 equals 𝑢 times Δ𝑡 squared
plus half times 𝑎 times Δ𝑡 squared. (C) 𝑠 equals half times 𝑢 times
Δ𝑡 plus 𝑎 times Δ𝑡 squared. (D) 𝑠 equals half times 𝑢 times
Δ𝑡 plus half times 𝑎 times Δ𝑡 squared.
Okay, so in this last bit of the
question, we’re given four different potential formulas for displacement 𝑠 written
in terms of the change in time Δ𝑡, the acceleration 𝑎, and the initial velocity
𝑢. Our task here is to work out which
of these formulas is the correct one. Recall that when we were answering
the first part of the question, we remarked that if we had an object whose motion
was represented on a velocity–time graph, then that object’s displacement was given
by the area under that velocity–time graph.
For the object in this question, it
starts at a velocity of 𝑢 at a time of 𝑡 one and ends up at a velocity of 𝑣 at a
time of 𝑡 two. That is, this pink line that we’ve
added is the line on the velocity–time graph which represents the object’s
motion. The area under this graph is all of
the area contained below this line between the times of 𝑡 one and 𝑡 two, so that’s
the area 𝐵 plus the area 𝐴. Since we know that this total area
under the graph represents the displacement of the object, then we can say that this
displacement is equal to area 𝐴 plus area 𝐵. Labeling this displacement as 𝑠
and recalling the results that we found for the areas 𝐴 and 𝐵, we then have that
𝑠 is equal to a half times 𝑣 minus 𝑢 times Δ𝑡 plus 𝑢 times Δ𝑡.
We want to find a way to rewrite
this equation so that it contains the quantity 𝑎, the acceleration of the
object. To do this, we can recall that the
acceleration of an object is defined as the rate of change of that object’s velocity
over time. That is, the acceleration 𝑎 is
equal to the change in the velocity of the object divided by the change in time over
which that velocity change occurs. In this case, since the object goes
from an initial velocity of 𝑢 to a final velocity of 𝑣, then its change in
velocity is equal to 𝑣 minus 𝑢. This velocity change takes place
between a time of 𝑡 one and a time of 𝑡 two. So, the change in time is equal to
𝑡 two minus 𝑡 one, which we’re told is equal to Δ𝑡. So, we can say then that the
acceleration 𝑎 is equal to 𝑣 minus 𝑢 all divided by Δ𝑡.
If we multiply both sides of this
expression by Δ𝑡, then on the right the Δ𝑡 in the numerator cancels with the Δ𝑡
in the denominator. This leaves us with an equation
that says Δ𝑡 multiplied by 𝑎 is equal to 𝑣 minus 𝑢. If we now look at this expression
we’ve got for the displacement 𝑠, we can see that in the first term on the right,
there’s this factor of 𝑣 minus 𝑢. But from our definition of
acceleration, we have found that 𝑣 minus 𝑢 is equal to Δ𝑡 multiplied by 𝑎. That means that we can replace this
𝑣 minus 𝑢 in the displacement equation by Δ𝑡 multiplied by 𝑎. So, in this expression for 𝑠, the
first term on the right-hand side is now a half times Δ𝑡 times 𝑎 times Δ𝑡.
We can rewrite that by grouping
together the two Δ𝑡 terms as a Δ𝑡 squared. So, we have that 𝑠 is equal to a
half times 𝑎 times Δ𝑡 squared plus 𝑢 times Δ𝑡. Finally, if we then swap the order
in which we write the terms on the right-hand side, we have that 𝑠 is equal to 𝑢
times Δ𝑡 plus half times 𝑎 times Δ𝑡 squared. This expression matches the formula
given in option (A). The formula which correctly
expresses the relationship between acceleration, displacement, and change in time
shown by the graph is 𝑠 equals 𝑢 times Δ𝑡 plus half times 𝑎 times Δ𝑡
squared.
