- #1

late347

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## Homework Statement

acceleratin as function of time

##a(t)= 2t+1##

we know that v(0)=0

and s(0)=0

find v(t)

find v(5)

find s(t)

find s(3)

and I was thinking about also what happens when t is negative number,

is it possible to find also v(-2)?

what about s(-3)?

## Homework Equations

integration rules, I suppose

## The Attempt at a Solution

I did the problem by

antidifferentiating the acceleration function

## a(t) = v'(t)##

## \int_{}^{} a(t) dt = v(t) ##

## v(t) = t^2 +t +c_{1}##

it was given that v(0)=0

now we can plug in v=0 and deduce that ##c_{1}## =0

v(5)=30for the distance as function of time s(t)

##s(t)= \int_{}^{} v(t) dt##

## s(t) =0.333...*t^3+ 0.5*t^2 +c_{2}##

it was given that s(0) =0

hence deduce that ##c_{2}=0##

s(3)=13.5 but I was little bit confused when my teacher's solution showed simply that

##v(t)= \int_{0}^{t} a(t)dt ##

I was confused about three things essentially

1.)why the lowerbound for definite integral has to be 0 as it was in my teachers solution

2.)why the upperbound is variable t, it seems like there's so many things happening in my teachers solution so I got a bit confused. I think the integral function notation is much clearer in my opinion.

3.)and what happens if I want to calculate v(-2), because... why not? what will be the upperbound and nthe lowerbound if I want to use my teacher's technique of definite integral?

OK I Know intuitively that when you have acceleration as function of time,

then the velocity has to be the integral of some sort ( dt is the width of a small rectangle, and multiply it by the a(t) the height to get the thin rectangle area, then you do the Riemann sum, velocity is the summed area)

It seems as though I somehow got the correct and same answer as my teacher but I'm still little bit confused about the whole thing as I described above.