Question

Using integral calculus:

A particle moving along a straight line increases its velocity by 2m/sec^2. If after 2 second its velocity is 5m/sec. and it has traveled 7m., find the distance traveled in the first 6 seconds.

Thanks! Step by step solution is much appreciated.

Answer 1

Alright I just started teaching my self integration recently and I have never done a problem like this, but if I am wrong I am sure someone else will come along and say something. I think you would ...
1. a=2x <--acceleration y''
2.Take the integral to get v=(1/2)x^2 +c <---velocity y'
3. Solve for c by putting in v(2)= 5
4. Take the integral of v=(1/2)x^2 +3 (because c ends up being 3)
5. Solve for c in d=(1/6)x^3 + 3x + c(distance y) by putting in d(2) = 7 to get c = -1/3
6 Then plug in d(6) to get the distance travelled in the first 6 seconds.

If I am correct the answer should be d(6) = 53.7m

Answer 2

first I'm gonna use the integral calc then use a physics formula to verify:
so our acceleration is constant therefore y''(x)=2, where y(x) is the position function or distance function which shows the distance travelled at a point. now integrate this to first to get y'(x)=2x+C. now we find the constant of integration by plugging y'(2)=5 5=2x+C C=1. so our velocity function is y'(x)=2x+1. integrating this we'll get y(x)=x^2+x+c. plug in y(2)=7 to get the c of this so 7=4+2+c c=1. so we have y(x)=x^2+x+1. so in the first 6 seconds, distance travelled is f(6)-f(0)=43+1=44.

Answer 3

Yeah I don't know what I was thinking acceleration is constant so obviously a=2 not 2x my bad looks like I had the right idea otherwise.

Answer 4

lol and I put (1/2) for no reason.

Answer 5

@anonymoustwo44 Thanks. I think I understand it now. I tried to verify it by kinematic equations but I got 45m. Waiting for the physics part. :)

Answer 6

@mathdood Thanks for replying as well :)

Answer 7

anonymous, I'm wondering about the part f(6)-f(0)=43+1=44. In my book, it says \[s(t_{2})-s(t_{1})=\int\limits_{t1}^{t2}v(t)dt\].

Shouldn't it be \[\int\limits_{0}^{6} (2t+1)dt\] ?

Answer 8

I found the answer using kinematic equations. here are my variables vf =13 , vi = 1, a= 2, d=? t=6 and here is the equation d = vi*t + (1/2)at^2 and it came out to be 42m. I was trying to figure out why my answer was wrong this time but I realized anon wrote f(6)-f(0)=43+1 when they meant to write 43-1 so the answer should have been 42 not 44. Now I feel less bad for my mistake haha =) even though it was worse because it was at the beginning of the problem instead of the end.

Answer 9

Also I think what you just wrote essentially means the same as what is in your book doesn't it? t2=6, t1=0, and v(t)=2t+1. This is just 43-1 =(6^2+6+1)-(0^2+0+1) right?

Answer 10

I have only glanced at definite integrals before and I didn't know how to do integrals at the time I just looked it up now to refresh my memory so once again there is a good chance I made a mistake. Let me know if I am understanding you marco26.

Answer 11

Yes, that's right. The answer is 42m. Thanks!

Answer 12

No thank you I felt like an idiot all day for making the 2x instead of 2 mistake. Now I feel I have redeemed myself haha.

Answer 13

oh sorry, I said " distance travelled is f(6)-f(0)=43+1. " what I did was f(6)+f(0) here so I should have done f(6)-f(0)=43-1=42