Hey!
Can you please help me with this question?
A particle moves according to x = t2 −8t+7, in units of metres and seconds. What is the maximum distance from the origin, and when does it occur: (i) during the first 2 seconds, (ii) during the first 6 seconds, (iii) during the first 10 seconds?
Hey!
So, we have
and we want to find maximum distance (ie. x). It's helpful to have a sketch in mind before attempting questions like this. Clearly it's a positive parabola. So, it doesn't actually have a 'maximum' turning point; rather it has a minimum, which we're not interested in. In fact, if you sketch the graph itself, you can get the answer seriously quickly.
Based on the above (or just intuition), it's clear that the maximum displacement will either occur at the beginning (ie. t=0) or end (ie t=2). We can just sub our values in to see which is correct!
It doesn't matter if the value is positive or negative, as long as it has the greatest magnitude. However, clearly the greatest displacement in the first section (first two seconds) occurs at t=0, x=7m.
Now, onto the first six seconds. This contains the global minimum, which may have a greater magnitude than x=7. So, we need to find it! We can use calculus for this part.
Subbing this value into our formula
So, clearly the point of greatest displacement is at t=4, x=-9.
Finally, we look to the first 10 seconds. The only way this any value could be greater than the global minimum is if it is far enough to the right that it keeps rising, and rising, and rising. So, we only need to check t=10.
Which is way bigger than -9m. So, the greatest displacement occurs at t=10, x=27m.