The acceleration (m/sec^2) of an object along a straight line is given in the attachment. Assuming that the object has an initial velocity of 5 m/sec , how far does the particle move during the time 0 ≤ t ≤ 3?

Question

anonymous · Answer

attachment

anonymous · Answer

pythagorean theorem.
3^2+10^2=d^2 (d as in distance)

auctoratrox · Answer

I think you have to find the linear function for acceleration and then take its integrals

auctoratrox · Answer

can anyone back me up on this

anonymous · Answer

I think you're looking into this too much bud. All the question is asking for is the distance that it travelled.

auctoratrox · Answer

but wouldn't the distance be gotten by a position function?

anonymous · Answer

Slope of the Acceleration: 3
Equation of Acceleration: $$y''=3x+1$$
Equation of Rate of Change $$y'=3x^2/2+x+5$$
Equation $$y=x^3/2+x^2/2+5x+C$$
since the graph is always increasing when x>0
we should get y(3)-y(0)
$$y(3) = 27/2+9/2+15 = 33$$
$$y(0) = 0$$
$$33$$

Does this look correct?

anonymous · Answer

Cxs, I solved your equation, and got 10.4 -- not on the answer list. What was your reasoning behind it? To the other that have answered: I guess I hadn't thought of doing that. I thought that by integrating twice I'd not be able to get the constant...

anonymous · Answer

Showing graph of acceleration not distance. My answer is not correct.

anonymous · Answer

Oh nightwill, that is great Thank you!

auctoratrox · Answer

yep. you got it

anonymous · Answer

Why integrate twice?

anonymous · Answer

Because I thought I'd have to do that to first get the v(t) function from acceleration and then again to get the position function.

anonymous · Answer

acceleration-->velocity-->distance..okay makes sense.