# Video: Pack 3 • Paper 3 • Question 20

Pack 3 • Paper 3 • Question 20

06:51

### Video Transcript

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 distance travelled.

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 two.

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 two, 11.

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 second.

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 55.8 meters.