Michella was practising her
sprinting. The graph shows the first 12
seconds of one particular sprint. Part a) Between which two times was
her acceleration at its peak? Part b) Work out the distance
travelled in her last two seconds.
For point a, we’re looking to find
the time during which her acceleration is highest. Another way to think about
acceleration is as change in a velocity over a given period of time.
On our graph, change in velocity is
change in the 𝑦-direction. And the change in time is change in
𝑥-direction. We can say then that the
acceleration is given by the gradient of the line. Remember gradient is change in 𝑦
over change in 𝑥.
This means we’re looking to find
the time during which the gradient is highest or the graph is steepest. We can observe that that’s this
part of the line. If we draw a vertical line down to
the horizontal axis, we see that that finishes at two seconds.
And we can, therefore, say that her
acceleration was at its peak between zero seconds and two seconds.
For part b, we’re looking to find
the distance travelled in the last two seconds of her journey. The graph finishes at 12
seconds. So the last two seconds are between
10 and 12 seconds.
And just like the gradient of a
velocity-time graph represents the acceleration, the area underneath it represents
the distance travelled. We can join these two parallel
lines and we see that we form a trapezium.
The formula for area of a trapezium
is 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 distance between them. We can spot quite easily that the
distance between the two parallel sides is two units.
But before we can work out the
lengths of the parallel sides themselves, we need to work out what the scale is on
this 𝑦-axis. Five small squares on the 𝑦-axis
represents two metres per second. And dividing through by five tells
us that one small square is equal to two-fifths of a metre per second or 0.4.
The longer parallel side ends two
squares above the number six. So that’s 6.8 units. And the shorter parallel side ends
two squares about four. So that’s 4.8 units. So we substitute these values into
our formula for the area of a trapezium. And we get one-half multiplied by
6.5 plus 4.8 multiplied by two.
Remember multiplication is
commutative. That means it can be done in any
order. If we do a half multiplied by two
first, those cancel each other out since one-half multiplied by two or one-half of
two is one. And this means the area of our
trapezium is simply 6.8 plus 4.8.
We can use a column method to work
this out. Eight plus eight is 16. So we put a six in the tenths
column and carry the one. We add a decimal point in the
answer directly below the decimal point in the question. Then, six plus four plus one is
11. This means the area of our
trapezium is 11.6 square units.
We can see that the velocity is
given to us in metres per second and the time is given to us in seconds. So the units for our distance are
The distance travelled in the last
two seconds of her journey was 11.6 metres.