Question

how to get acceleration give the velocity and v(t)

Answer 1

and v = 16.0m/s

Answer 2

ohh sorry, v(t) = (0.860m/s^3)t^2

Answer 3

so what is the derivative of v(t),
what is v'(t)?

Answer 4

v'(t) = (2)0.860t

Answer 5

right and you are evaluating at t=16.0 m/s

Answer 6

how come t = 16.0m/s? isn't 'm/s' a unit for velocity?

Answer 7

whops , your right, my mistake

Answer 8

haha..no problem :)

but can I get the time by equating it v'(t) to the value of velocity (which is 16.0m/s)?

Answer 9

but the unit I get for the time by doing that is 's2'.

is that allowed?

Answer 10

can i see your working

Answer 11

\[since v'(t) = (2)(0.860 m.s^3)t = 16.0m/s\]

Answer 12

your right about the units being wrong

Answer 13

I think you have to find the time that the velocity is 16 m/s
with the original equation first.

find \(t\)
\[v(t)=0.860 t^2 [\text m/\text s^3] = 16 [\text m/\text s]\]

Answer 14

and sub this \(t\) into \(v'(t)\) that you already found

Answer 15

ohhh. I think that's possible :)

Answer 16

after getting t...where I should I substitute it to get acceleration?

Answer 17

\[a(t)=v'(t) = (2)0.860t\]

Answer 18

sorry about confusing you earlier

i can check your \(t\) and \(a(t)\) if you want

Answer 19

Ohh. Okay so I tried equating v'(t) to v.

Answer 20

yeah that wont work, because v' is acceleration

Answer 21

Then I get t=4.3133 s

So, should I substitute the value of t to v'(t) to get acceleration?

Answer 22

that's better, yes

Answer 23

I think I got it already :) a(t) = 7.4189 m/\[s ^{2}\]

Answer 24

\[\checkmark\]

Answer 25

Thank you so much sir! :) God Bless! ^^

Answer 26

Good work.