Question

im not sure how this is related to derivatives at LL

Answer 1

Derivatives are not needed for questions 15 thru 19 that you have posted.

Answer 2

@kropot
how do you do number 15?

Answer 3

@kropot72

Answer 4

It is related to derivatives in this is a v-t graph.

eg it stops climbing and so maxes out when v = 0, ie t = 8

cf your answer to #17

look at the v-t profile to see the different types of acceleration.

eg constant \(\large \frac{\Delta v}{\Delta t} = g\) when it's just gravity acting...

Answer 5

and for displacements you're going to have to count the little squares under the v-t curve

= integration

Answer 6

and how is that related to derivative? @IrishBoy123

Answer 7

To answer number 15, you just need to interpret the graph correctly. What is the greatest value of velocity reached and how many seconds after launch did this value of velocity occur?
Note: Your answers to number 17 are not correct. As @IrishBoy123 posted: "it stops climbing and so maxes out when v = 0, ie t = 8".

Answer 8

in these circumstances:

slope = derivative

\(a = \dfrac{dv}{dt}\)

Answer 9

For number 15, you are asked for the maximum upward velocity. This can be read off the scale on the vertical axis, without doing any calculation. Hint: The maximum velocity is reached about 2.1 seconds after launching.