Help with Calc/Physics Question - Integration Methods

[Chriss]

1. The Problem
The acceleration function for a particle is given as a(t)=60t+18. v(1)=5, find the distance traveled by the particle from t=0 to t=2

The Attempt at a Solution

First, I integrated the acceleration function to get the velocity function, I found:

v(t)=30t²+18t+c
v(1)=5=30+18+C
5=48+C
C=-43
So v(t)=30t²+18t-43

Well, now I scratched my head a little bit. I know that the distance traveled is the absolute value of the integral of this function on my interval, but with the -43 hanging out on the end of this equation, I will not be able to factor. I approached my prof. about this and he said there was a formula that I should use, and that the quadratic equation solution that I produced wasn't necessary, and that this problem could be done on a scientific calculator.

What am I missing?

(I had hoped that v(1) would be 0 so it would split right down the middle )

TIA for the help.

Chris

 

[SteamKing]

Well, I'm scratching my head, too. You apparently know that integrating the acceleration w.r.t. time will give velocity. How is velocity related to knowing distance and time? From what I recall of rectilinear motion, the answer doesn't involve the quadratic formula.

 

[Chriss]

If you find the net area under the velocity curve (absolute value), then you end up with your distance traveled.

To determine where the velocity curve may change from positive to negative, we need to know where velocity is zero. This would allow one to create two separate integrals to find the absolute value of the area under the curve.

 

[SteamKing]

I'm still scratching my head. The OP just wants the distance traveled from t = 0 to t = 2. It doesn't say distance traveled w.r.t. some fixed point, just distance travelled. I think you are overthinking this problem.

 

[vela]

1. The Problem
The acceleration function for a particle is given as a(t)=60t+18. v(1)=5, find the distance traveled by the particle from t=0 to t=2

The Attempt at a Solution

First, I integrated the acceleration function to get the velocity function, I found:

v(t)=30t²+18t+c
v(1)=5=30+18+C
5=48+C
C=-43
So v(t)=30t²+18t-43

Well, now I scratched my head a little bit. I know that the distance traveled is the absolute value of the integral of this function on my interval, but with the -43 hanging out on the end of this equation, I will not be able to factor. I approached my prof. about this and he said there was a formula that I should use, and that the quadratic equation solution that I produced wasn't necessary, and that this problem could be done on a scientific calculator.

What am I missing?

(I had hoped that v(1) would be 0 so it would split right down the middle )

TIA for the help.

Chris

I don't think you're missing anything. I'm curious to know what your professor has in mind.

 

[Chestermiller]

What makes you feel like you have to factor something?

Chet

 

[SammyS]

Let's see:

The particle leaves its initial position, which we can assume is the origin, then travels in the negative direction to position x₁ where it changes direction passing the origin and arrives at position x₂ 2 seconds after leaving the origin. The distance traveled is arrived at quite easily from this.

Using a graphing calculator and the position function, one can easily answer the question being asked here. Perhaps that's what the prof. has in mind.