Find the distance.

Question

Find the distance.

nnesha · Answer

|dw:1461668102830:dw| this is what i got for the first part (a) $$\frac{t^2}{2}-6t+C$$ $$10=C$$$$y=t^2-6t+10$$ i'm not getting the correct answer for b) http://image.prntscr.com/image/9d75a62e72d9456b87da832818543f5f.png $$\int\limits_{0}^{3} t^2-6t+10 +\int\limits_{3}^{4} (t^2-6t+10)-(2t-6)$$ is this correct ??

nnesha · Answer

|dw:1461668878486:dw|

rvc · Answer

v(t)=2t-6
$$\rm \color{red}{2}\frac{t^2}{2}-6t+C$$

rvc · Answer

initial position 
so put t=0
to find C

rvc · Answer

now we need to find the distance

nnesha · Answer

the anti derivative of 2t -6 is $$\frac{ \cancel{2}t^2 }{ \cancel{2} } -6t+C$$ ?

rvc · Answer

|dw:1461670529551:dw|

rvc · Answer

draw this 
i hope it should help
because i used to do the same XD

nnesha · Answer

draw what ?

rvc · Answer

the curve
velocity vs time
and 
distance vs time

nnesha · Answer

$$\rm y=t^2-6t+10$$ this is correct 
i just need help with b) :(

rvc · Answer

im lazy enough to search my mech book XD

nnesha · Answer

aww ;-; amiiii g :(

rvc · Answer

wait a moment

nnesha · Answer

ok thanks<3
btw i have to go in like 15 minutes

rvc · Answer

is it 0th sec to 5th sec?

rvc · Answer

wait wait 
it is not written so ..

nnesha · Answer

|dw:1461671467907:dw| yep :)