Question

Apebblefallsfromthetopofacliffthatis180mhigh.Thepebble’sheight above the ground is modelled by h(t) 􏰀 􏰂5t2 􏰂 5t 􏰁 180, where h is the height in metres at t seconds since the pebble started to fall.
(a) Find the average rate of change of height between 1 s and 4 s.
(b) Find h(3).
(c) Find the instantaneous rate of change of height at 3 s.
(d) Communication: Explain the meaning of each value you calculated for (a), (b), and (c).

Answer 1

Ah the first part of your problem is jumbled up

Answer 2

h(t) = -5t2 - 5t + 180,

Answer 3

Average rate of change between two points is same as the slope of segment joining those points

Answer 4

How can I do it? Please help me

Answer 5

Average rate of change between t = 1 and t = 4 :
\[\rm \dfrac{h(4) - h(1)}{4-1}\]

Answer 6

evaluate the given height function at those values and plugin

Answer 7

Is it -30?

Answer 8

Correct!

Answer 9

Wow! Thanks!
Is my answer in letter B correct? It's 120
How about letter C?

Answer 10

120 is right

Answer 11

are you familiar with derivatives ?

Answer 12

No

Answer 13

you need to find derivative of h(t) for instantaneous rate

Answer 14

How will I do it? Please please help me.

Answer 15

There are many ways to work part c, im just not sure which way your teacher wants you work it

Answer 16

this question is from calculus right ?

Answer 17

i hope its not from precalculus

Answer 18

Yes

Answer 19

then you should be knowing differentiation

Answer 20

This is our 2nd lesson I really don't know differentiation

Answer 21

Ahhh! Is it like the limit as x2 is approaching x1 ?

Answer 22

you know limits ?

Answer 23

then for part c, you just need to find below limit :

\[\rm \text{instantaneous rate of change at 3s = } \lim\limits_{t\to 3}\dfrac{h(t) - h(3)}{t-3}\]

Answer 24

Yes

Answer 25

\[\rm = \lim\limits_{t\to 3}\dfrac{ -5t^2 - 5t + 180 - 120}{t-3}\]

Answer 26

\[\rm = \lim\limits_{t\to 3}\dfrac{ -5t^2 - 5t + 60}{t-3}\]

Answer 27

\[\rm = \lim\limits_{t\to 3}\dfrac{ -5(t^2 + t - 12)}{t-3}\]

Answer 28

\[\rm = \lim\limits_{t\to 3}\dfrac{ -5(t+4)(t-3)}{t-3}\]

Answer 29

\[\rm = \lim\limits_{t\to 3} -5(t+4)\]

Answer 30

\[\rm = -5(3+4)\]

Answer 31

\[\rm = -35\]

Answer 32

thats the instantaneous rate of change of height at t = 3s

Answer 33

But can I also use the table ?

Answer 34

Thankyou again :)
How can I explain it on letter D?

Answer 35

Average rate of change of height is same as `velocity`

Answer 36

Average rate of change of height betwee two points is same as the `average velocity` between those points

Answer 37

Since you got -30 for part a, that means the average velocity of the pebble is -30 m/s between 1s and 4s

Answer 38

How about letters b and d?

Answer 39

I mean c

Answer 40

whats your interpretation for part b ?

Answer 41

how do u interpret h(3) ?

Answer 42

I think in 3s time the pebble is in the height 120m

Answer 43

Very good!

Answer 44

for part c :
the velocity of particle at t=3s is -35 m/s

Answer 45

Yayyyy! I super thank you, :D

Answer 46

yw:)

Answer 47

How can I give you a medal? :)

Answer 48

looksyou already gave me one :) you need to click on "Best response" button next to a reply

Answer 49

Yah I already did it, Thankyou again :)