Question

The position of a point on a line is given by the equations(t)=t^3−6t^2+9t−4 , where s is measured in metres and t in seconds.

What is the velocity of the point after 2 seconds?

What is its acceleration after 4 seconds?

Where is it when is first stops moving?

How far has it travelled when its acceleration is 0?

After 2 seconds, is it moving toward or away from the origin?

Answer 1

@ashhhrodriguez @appleduardo @bohotness @ganeshie8 @GodGirl360 @nelsonjedi

Answer 2

I tried it 3 different ways still cant do it help plz

Answer 3

Velocity is the first derivative
Acceleration is the second derivative.

Answer 4

are u sure coz it says the equation gives position and not displacement

Answer 5

velocity and acceleration are derivatives of position

Answer 6

so the first derivative is velocity and 2nd derivative is acceleration?

Answer 7

Wish i could help you right now @Wazzabob, but I've got three tests this week(one of them being a final) and I need to get my homework done. Although, I would suggest putting the formula in on a graphing calculator(you can find plenty online). Seeing the line may help you solve your questions. Good luck!

Answer 8

Yes. Then plug in your t value to determine your acceleration and velocity.

Answer 9

So what is the formula for velocity going to be>

Answer 10

Use the power rule

Answer 11

1st derivative :
3t2−12t+9

2nd Derivative : 6t-12
3 derivative : 6

Answer 12

i have worked it out completely, i send you my answers

Answer 13

Good so the velocity at 2 seconds will be?

Answer 14

im not sure if i got the correct answer for part d)

Answer 15

Part d make acceleration =0, Thus 0=6t-12. Looks good to me. Good job

Answer 16

so is it right, is distance travelled =2m when acceleration =0?

Answer 17

esY

Answer 18

hey thank you very much @nelsonjedi this assignment counts for my final thats why i needed a second opinion thanks again :)

Answer 19

Anytime..good luck