# acceleration/velocity

• Aug 9th 2009, 09:56 AM
skeske1234
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?
• Aug 9th 2009, 10:19 AM
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?

http://www.ballooncalculus.org/asy/diffPower/diff.png

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

... and did you differentiate again?

http://www.ballooncalculus.org/asy/diffPower/diff2.png

(to show a(t) is constant)
______________________________________

Don't integrate - balloontegrate!

Balloon Calculus Forum
• Aug 9th 2009, 10:45 AM
skeske1234
Quote:

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?

http://www.ballooncalculus.org/asy/diffPower/diff.png

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

... and did you differentiate again?

http://www.ballooncalculus.org/asy/diffPower/diff2.png

(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
• Aug 9th 2009, 10:59 AM
tom@ballooncalculus
http://www.ballooncalculus.org/asy/diffPower/diff3.png

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?
• Aug 9th 2009, 11:26 AM
skeske1234
Quote:

Originally Posted by tom@ballooncalculus
http://www.ballooncalculus.org/asy/diffPower/diff3.png

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?
• Aug 9th 2009, 11:29 AM
tom@ballooncalculus
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).