Falling body problem

**Lambic** · November 18th 2009, 07:56 AM

Given the acceleration function A(t)= 12t^2 + 2

A] Find the Velocity function V(t) with the initial condition when t = 0 then V = 7

B] Find the Distance function S(t) with the initial condition when t = 0 then S = 12

Not sure where to start or if someone could help walk me through this.

**tonio** · November 18th 2009, 08:13 AM

Originally Posted by Lambic

Given the acceleration function A(t)= 12t^2 + 2

A] Find the Velocity function V(t) with the initial condition when t = 0 then V = 7

Ac acceleration is the velocity function's derivative, here you have to antiderivate (or evaluate the indefinite integral) of the acceleration, and find out what the inegral's constant is according to the given initial conditions:

$V(t)=\int(12t^2+2)dt=4t^3+2t+C=$ constant. But we're said $t=0\Longrightarrow\,7=V(0)=C\Longrightarrow\,V(T)= 4t^3+2t+7$

B] Find the Distance function S(t) with the initial condition when t = 0 then S = 12

As the velocity is the distace function's derivative, again here you have to evaluate the integral of the velocity and find out what the integral constant is according to the given initial condition, just as we did above.

Tonio

Not sure where to start or if someone could help walk me through this.

.

**Amer** · November 18th 2009, 08:15 AM

Originally Posted by Lambic

Given the acceleration function A(t)= 12t^2 + 2

A] Find the Velocity function V(t) with the initial condition when t = 0 then V = 7

B] Find the Distance function S(t) with the initial condition when t = 0 then S = 12

Not sure where to start or if someone could help walk me through this.

$v = \int A \cdot dt$

$v(t) = \int 12t^2 +2 \cdot dt = 4t^3 + 2t + C$

but when t=0 v=6

$v(0) = 4(0) + 2(0) + C =6 \Rightarrow C=6$

$S(t) = \int v \cdot dt$

$S(t) = \int 4t^3 + 2t +6 \cdot dt = t^4 + t^2 + 6t +C$

but when t=0 S=12

$S(0) = 0+0+0+C =12 \Rightarrow C=12$ so you have

$v(t) = 4t^3 + 2t +6$

$S(t) = t^4 +t^2 +6t+12$

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