Derive an integration equation

[clobber]

The acceleration, a, of an object is given by a = 3t² - 6t, where t is the time elapsed, measured in seconds. Derive an equation for the velocity of the object, v, in terms of t.

So in integration you increase the power and divide by the new power. so doing that using the above info i get this

3t³/3 -6²/2

=t³-3²

is this right?

 

[skeeter]

you forgot the t (which is squared, not the 6) in the second term and the constant of integration

 

[clobber]

Thank you

 

[Deleted member 4993]

The acceleration, a, of an object is given by a = 3t² - 6t, where t is the time elapsed, measured in seconds. Derive an equation for the velocity of the object, v, in terms of t.

So in integration you increase the power and divide by the new power. so doing that using the above info i get this

3t³/3 -6²/2

=t³-3²

is this right?......................NO

NO!

1. \(\displaystyle \frac{6^2}{2}\) is NOT equal to 3^2 .............. it should be\(\displaystyle \frac{6}{2}\) instead of \(\displaystyle \frac{6^2}{2}\)..................................edited
   
   
2. ............ and multiply by 't²' making it\(\displaystyle \frac{6t^2}{2}\).......... fix it..................................edited
   
   
3. You missed the constant of integration ....................... please add 'C' to the expression of v(t)

 

[JeffM]

NO!

1. \(\displaystyle \frac{6^2}{2}\) is NOT equal to 3^2 ........................ it should be 18 ............ and multiply by 't²'.......... fix it
   
   
2. You missed the constant of integration ....................... please add 'C' to the expression of v(t)

Before Jomo sees this, [MATH]\dfrac{6t^2}{2} = 3t^2 \ne 18t^2.[/MATH]
Go to the corner for 3 minutes. Jomo might sentence you to 3 MONTHS, but now you can plead double jeopardy.