1. ## acceleration/velocity

The distance-time relationship for a moving object is given by s(t)=kt^2+(6k^2-10k)t+2k, where k is a non-zero constant.

a) show that the acceleration is constant

b) find the time at which the velocity is zero and determine the position of the object when this occurs.

a) i cant seem to get the derivative of this correct.

b) v(t)=0 solve for t. how do i figure out the position?

2. Are you clear that v(t) is the derivative of s(t), and acceleration the derivative of that? If so, then your only problem is getting the derivatives right. For the first, can you see that the blank here will just be the bracketed expression involving k?

(set v(t) to zero and solve for t)

... and did you differentiate again?

(to show a(t) is constant)
______________________________________

Don't integrate - balloontegrate!

Balloon Calculus Forum

3. Originally Posted by tom@ballooncalculus
Are you clear that v(t) is the derivative of s(t), and acceleration the derivative of that? If so, then your only problem is getting the derivatives right. For the first, can you see that the blank here will just be the bracketed expression involving k?

(set v(t) to zero and solve for t)

... and did you differentiate again?

(to show a(t) is constant)
______________________________________

Don't integrate - balloontegrate!

Balloon Calculus Forum
yes.. the problem is that i am not sure if i have my differnetiation right... so can you continue with that?

so far i have: s'(t)=14kt +6k^2-10t-10k+2
s''(t)=26k-18
not sure now what to do

4. Are you clear that v(t) = s'(t)?

And that k is just a constant, so that e.g. $\displaystyle 6k^2 - 10k$ is also just a constant?

5. Originally Posted by tom@ballooncalculus

Are you clear that v(t) = s'(t)?

And that k is just a constant, so that e.g. $\displaystyle 6k^2 - 10k$ is also just a constant?
constant meaning derivative is 0 right?

6. Yup, if the constant is the whole term. Whereas $\displaystyle kt^2$ becomes $\displaystyle 2kt$ because as the multiplier ('coefficient') of a term a constant gets (during differentiation) multiplied by the current power (which then in turn gets reduced by one).