Question

can someone please please please explain this calc problem for me??

Answer 1

only number 2 !

Answer 2

What did you get for 1?

Answer 3

10/3

Answer 4

You noticed that it's the average velocity that was needed, which is
average velocity = change in displacement / time

Answer 5

yeah..

Answer 6

am i wrong?

Answer 7

or
average velocity = (final position-initial position) / time

Answer 8

was the first one wrong?

Answer 9

No, it was my bad.
10/3 is right, I went too fast with v(t) instead of finding s(t).
So what's bugging you for #2?

Answer 10

can you explain it?

Answer 11

for #2?

Answer 12

yess

Answer 13

like step by step please

Answer 14

For #2, hint:
velocity is signed. Speed is not. That's the only tricky part with #2.
Can you figure it out?

Answer 15

ok can we break it down... thats why i posted it...

Answer 16

so the first step is to find the derivative?

Answer 17

Have you found the instantaneous velocity at t=5?

Answer 18

2t-9

Answer 19

Given v(t)=\(t^2-9*t+18\)

Answer 20

ok idk what youre doing lol

Answer 21

2t-9 would be dv(t)/dt = acceleration.
The question needs the instantaneous velocity!

Answer 22

i just plug in 5?

Answer 23

Exactly! :)

Answer 24

no derivative?

Answer 25

Nope, not according to the question.

Answer 26

-2?

Answer 27

Yes for velocity!

Answer 28

and now for speed?

Answer 29

Speed is 2m/s in whichever direction, basically it's the magnitude of the velocity.

Answer 30

wait how did you get that?

Answer 31

Velocity = -2 m/s, which is towards the "left".
For speed, you just drop the sign to get 2 m/s, since it is always positive.

Answer 32

ohh ok thanks(:

Answer 33

Perhaps if you read up a little about the difference between speed and velocity, distance and displacement, it would help.
Teachers like to make tricky questions like this one.

Answer 34

You're welcome! :)