For an object whose velocity in ft/sec is given by v(t) = −3t^2 + 5, what is its displacement, in feet, on the interval t = 0 to t = 2 secs?

a: 6.607
b: 2
c: −2.303
d: 2.303

Question

For an object whose velocity in ft/sec is given by v(t) = −3t^2 + 5, what is its displacement, in feet, on the interval t = 0 to t = 2 secs?

 a: 6.607
 b: 2
 c: −2.303
 d: 2.303

anonymous · Answer

@misty1212 can u plz help? :)

anonymous · Answer

I think its a, 6.607 but double-checking is good :)

misty1212 · Answer

if i am not mistaken it is 
$$\int_0^2(-3t+5)dt$$

solomonzelman · Answer

no, this is correct.
s(t) is integral of v(t)

misty1212 · Answer

oh but i wrote it wrong doe

solomonzelman · Answer

yes, but conceptualized correctly:)

misty1212 · Answer

$$\int_0^2(-3t^2
+5)dt$$

misty1212 · Answer

your answer, however, is wrong

anonymous · Answer

kk 
integral...

solomonzelman · Answer

put it inline,      $\large\color{slate}{\displaystyle\int\limits_{0}^{2} \left( -3t^2+5\right)~~dt}$

solomonzelman · Answer

`$\large\color{slate}{\displaystyle\int\limits_{0}^{2}  \left( -3t^2+5\right)~~dt}$`
(don't mean to give a latex class)

anonymous · Answer

? i can't see it

misty1212 · Answer

that does make the numbers look better for sure

misty1212 · Answer

$$\int\limits_{0}^{2}\left(-3t^2+5\right)dt$$

misty1212 · Answer

thnx!

anonymous · Answer

thank u :) i can see that but do u mind walking me thru the integration process

anonymous · Answer

well i think its c, -2.303 @misty1212 :)