The graph below shows information
about the velocity, 𝑣 meters per second, of a football 𝑡 seconds after it begins
to roll down a hill. Work out an estimate for the
acceleration of the ball at 𝑡 equals two.
There are two pieces of information
we can work out from our velocity time graph. The first is that the gradient of
the graph represents the acceleration at that time. The second is that the area under
the curve — that’s the area between the curve and the 𝑥-axis — represents the total
To answer this question then, we’ll
need to calculate the gradient of the curve at 𝑡 equals two. Since the graph is a curve and not
a straight line, we’ll need to first construct the tangent to the curve at 𝑡 equals
The tangent is the line that
follows the general direction of the curve at this point. And it will need to be drawn by
eye. Once this tangent is drawn, we can
use the formula for gradient of a straight line: change in 𝑦 over change in 𝑥. That’s often written as 𝑦 two
minus 𝑦 one over 𝑥 two minus 𝑥 one, where 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two are
any two coordinates on the tangent to the curve.
Let’s choose two coordinates that
lie on this line. The first sensible choice would be
the point at which we’ve drawn the tangent. That’s when 𝑡 is equal to two. At this point, the coordinate is
The next choice could be anywhere
on the tangent. But we do need to be a little bit
careful with the scale on the 𝑦-axis. We can see that five little squares
represent one meter per second. If we divide these numbers by five,
it shows us that one little square represents one-fifth of a meter per second. That’s equivalent to 0.2 meters per
Let’s choose the point on the
tangent when 𝑡 is equal to one. Since this point is situated two
little squares above 12 on the 𝑦-axis, it must have a 𝑦-value of 12.4. The coordinate is one, 12.4.
Now that we have these two
coordinates, we decide which we want to be the first coordinate and which we want to
be the second. It doesn’t really matter which way
around we choose as long as we are consistent. Let’s call two, 11 the first
coordinate and one, 12.4 the second.
Substituting these values into our
formula for the gradient of a straight line gives us 12.4 minus 11 all over one
minus two, which is negative 1.4. The acceleration at 𝑡 equals two
seconds is, therefore, negative 1.4 meters per second squared. Another way of saying this is that
the ball is decelerating at a rate of 1.4 meters per second squared.
Remember the tangent to the curve
was drawn by eye. What this means then is it
sometimes the coordinates we get might be slightly different to the coordinates in
the mark scheme. In fact, there’ll usually be a
range for the gradient. In this case, it’s between negative
1.25 and negative 1.75.
Using three strips of equal width,
work out an estimate for the total distance the football has travelled nine seconds
after it is kicked.
We know that the area between the
curve and the 𝑥-axis tells us the total distance travelled. We can split the area up into
polygons to help us calculate an estimate for the total distance travelled. Since we’re told that the graph
should be split into three strips, we can begin by calculating the width of each
strip. The total width of the curve is
nine units. So we can divide this by three to
give us a width of three units for each strip.
We can also find the coordinates of
the points where each line intersects the curve. Since each little square on the
𝑦-axis represents 0.2 meters per second, they are zero, 12.4; three, 9.2; six,
3.2. And the final coordinate we’re
interested in is at nine, zero.
Next, let’s turn these strips into
polygons that we know how to find the area for. That’s two trapeziums and one
triangle. The formula for the area of a
trapezium is given by a half multiplied by 𝑎 plus 𝑏 multiplied by ℎ, where 𝑎 and
𝑏 are the lengths of the parallel sides and ℎ is the height between them.
Let’s substitute what we know about
the first trapezium into this formula. The lengths of the two parallel
sides are found by looking at the 𝑦-coordinates of the points where these sides
intersect the curve. That’s 12.4 and 9.2. The distance between these parallel
sides is three units. The area is then given by a half
multiplied by 12.4 plus 9.2 multiplied by three, which is 32.4 units squared.
Let’s repeat this process for the
second trapezium. The lengths of its parallel sides
are 9.2 and 3.2 units. And the distance between these
sides is three units. The formula then becomes a half
multiplied by 9.2 plus 3.2 multiplied by three, which is 18.6 units squared.
The third shape is a triangle. The formula for the area of a
triangle is a half multiplied by the length of its base multiplied by its
height. The length of the base of this
triangle is three units and its height is 3.2 units. A half multiplied by three
multiplied by 3.2 is 4.8 units squared. We must add these values together
to work out an estimate for the total area. That’s 32.4 plus 18.6 plus 4.8. An estimate for the total area then
is 55.8 units squared.
The area represents the total
distance travelled. And the original units for velocity
are in meters per second. The total distance travelled is