Question

Calc AB Question
I have an answer, I just need to know if my answer is correct.

The function v(t) = t^2 - 5t is the velocity in m/sec of a particle moving along the x-axis, where t is measured in seconds. Write an expression for the displacement and the total distance traveled by the particle on the interval 0
12 years ago

Answer 1

u just need to integrate this eq with limits 0 and 10

Answer 2

My answer is \[s(t) = \int\limits_{0}^{10}(t^2-5t)dt\]

Answer 3

All it says to do is write the expression, it doesn't say to solve.

Answer 4

s(t) is displacement right?

Answer 5

yes it gives u s(10) - s(1)

Answer 6

so clearly its a displacement

Answer 7

*s(10) - s(0)

Answer 8

Lol ok I'm just making sure.

Answer 9

for distance u need to take abs value of velocity function

Answer 10

Yes I know. @sourwing taught me how to do distance.

Answer 11

distance :

\(\large s(t) = -\int\limits_{0}^{5}(t^2-5t)dt + \int\limits_{5}^{10}(t^2-5t)dt \)

Answer 12

okay good :)

Answer 13

s(t) isn't displacement, it's position.

displacement = ∫ v(t) dt , from a to b

Answer 14

I thought x(t) was position

Answer 15

distance :

\(\large s(t) = \int |(t^2-5t)| dt \)

displacement :

\(\large s(t) = \int (t^2-5t)dt \)

Answer 16

for dispacement u can simply take take : s(10) - s(0)

Answer 17

those x and s are just letters they pick to represent the position. Certainly you could have easily chosen p(t) to represent the position

The reason I said s(t) isn't displacement is because it is position.

if s(5) = 5, this only means the position is 5 unit away from 0.

Displacement is change in position, so you will need two t value

Answer 18

^^

Answer 19

Ohhhhhhhhh lol (it clicked)

Answer 20

that's why
displacement = ∫ v(t) dt, from a to b

if you just do ∫v(t) dt without the limit of integration, then it is a position function

Answer 21

Oh true

Answer 22

Well thanks again @sourwing !! That's the last question for today haha