Question

@perl can you help me on this calc problem?

Answer 1

speeding up : a >0 & v > 0 or a < 0 & v < 0

slowing down: a < 0 & v > 0 or a > 0 & v < 0

Answer 2

wait whaat?

Answer 3

ohh

Answer 4

how do ifind the time interval??

Answer 5

@perl

Answer 6

one moment

Answer 7

okk

Answer 8

it is speeding up on (3, 4.5) ( 6, 8)

it is slowing down on (0, 3) , (4.5, 6)

Answer 9

how did you get that? im just wondering

Answer 10

you can take the derivative of v(t) to get acceleration

Answer 11

it is speeding up when velocity and acceleration have the same sign (both positive or both negative)

it is slowing down when velocity and acceleration have different sign

Answer 12

so the derivative is 2t-9, correct? @perl

Answer 13

can you write the steps please??

Answer 14

@perl ???

Answer 15

a(t) = 2t - 9 , when is this positive, when is it negative

Answer 16

when it is negative?

Answer 17

solve a(t) > 0 , a(t) < 0

Answer 18

2t - 9 > 0 when t > 9/2

2t - 9 < 0 when t < 9/2

Answer 19

ohh

Answer 20

4.5?

Answer 21

yes

Answer 22

oh ok so thats it

Answer 23

and we know what v(t) > 0

Answer 24

yeah

Answer 25

*when

Answer 26

okk

Answer 27

thanks(: