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

You know how to find the derivative?

Answer 2

yes, but im not sure if the first derivative is displacement or velocity coz the question says that the equation gives the position of the point, does the position means displacement?

Answer 3

1st derivative : \[3t^{2}-12t+9\]
2nd Derivative : 6t-12
3 derivative : 6

Answer 4

the equation as it is now is position over time (probably called 'displacement' but I'm not up with the math lingo)

take the derivative of what you posted, and that will be the equation for velocity over time I think..

Then if you take the derivative of the velocity equation you should get the acceleration equation.

Answer 5

I agree with your derivative math.

Answer 6

hey wait i second i post the answers i already got, I worked it out already just needed a second opinion

Answer 7

Sure thing -- the math looks good; though I'm not positive that the 1st derivative = velocity and the 2nd = acceleration- you might want a 3rd opinion.

Answer 8

for part b -- why is the final answer to the -2 power?

Answer 9

ms^-2 is same as m/s^2