Question

The velocity, at time t, of a point moving on a coordinate line is (1+t^2)^-1 ft/sec.
At what time, t, t>0, do the acceleration and velocity have the same absolute value?

Answer 1

I havent a clue how to do this

Answer 2

@Loser66 any ideas?

Answer 3

why me?

Answer 4

its AP calc practice

Answer 5

Because I have no idea yet.

Answer 6

hahaha. .. neither I ,

Answer 7

If you can explain it to me how to solve it I should be ok

Answer 8

ok, we can work together, a = v' and let them equal to solve for t, is it ok?

Answer 9

ok so a is the acceleration t is the time and v is the velocity

Answer 10

I got an idea guys.

Answer 11

that is

Answer 12

oh yeah i remember that acceleration is the second derivative and velocity is the first derivative

Answer 13

so acceleration would be the derivative of the velocity

Answer 14

I got t =1

Answer 15

I was eating ruffles. @Loser66

Answer 16

so I take the derivative of the (1+t^2)^-1 correct?

Answer 17

hahaha, good for you

Answer 18

@Valkarie70 yes

Answer 19

ok then I take the absolute value of that and it should equal (1+t^2)^-1 when t=1

Answer 20

yes, I think so,

Answer 21

Ok I will check

Answer 22

hey, where is @Isaiah.Feynman that guy eats ruffles and disappears, hahahaha

Answer 23

yes because (1+1^2) is 2 and 2^-1 is 0.5

Answer 24

and the derivative of the velocity calculated at x=1 is 0.5

Answer 25

THANK YOU :D

Answer 26

oh, so, I am so lucky when giving out the correct instruction, hehehe, thanks for medal

Answer 27

np :)

Answer 28

Right here! @Loser66 I will quickly if my idea produces your answer of t=1.

Answer 29

I don't understand your sentence when there is no verb in it "I will quickly if...."

Answer 30

I omitted the word "check"

Answer 31

hehehe.. ok got it now