antiderivative particle problem

[jbake77]

A particle is moving with the given data. Find the position of the particle.

a(t)=10+3t-3t^2, s(0)=0, s(2)=10

So, I started out by taking the antiderivative of a(t)
I got----> s(t)=10t+(3/2)t^2-t^3+c
Then I plugged in the S(0) and get c=0
so i plugged that into s(t)
next i plugged s(2)
I got----> s(2)=10(2)+(3/2)2^2-s^3+c and got c=-8
I got my final answer to be s(t)=10t+(3/2)t^2-t^3-8
not sure if that was correct if anyone could fix my errors!

 

[soroban]

Hello, jbake77!

A particle is moving with the given data. .Find the position of the particle.

. . a(t) .= .10 + 3t - 3t², . s(0) = 0, .s(2) = 10.

v(t) . = . ∫ a(t) dt . = . 10t + (3/2)t² - t³ + C₁

s(t) . = . ∫ v(t) dt . = . 5t² + (1/2)t³ - (1/4)t⁴ + C₁t + C₂


s(0) = 0: . 0 + 0 - 0 + 0 + C₂ .= .0 . . → . . C₂ = 0

s(2) = 10: . 20 + 4 - 4 + 2C₁ + 0 .= .10 . . → . . C₁ = -5


Therefore: . s(t) . = . -5t + 5t² - (1/2)t³ - (1/4)t⁴

 

[HallsofIvy]

A particle is moving with the given data. Find the position of the particle.

a(t)=10+3t-3t^2, s(0)=0, s(2)=10

So, I started out by taking the antiderivative of a(t)
I got----> s(t)=10t+(3/2)t^2-t^3+c

That would have correct if this were the speed. The acceleration is the second derivative, not the first. You will have to integrate again so that you will have TWO "constants of integration".

Then I plugged in the S(0) and get c=0
so i plugged that into s(t)
next i plugged s(2)
I got----> s(2)=10(2)

+(3/2)2^2-s^3+c and got c=-8
I got my final answer to be s(t)=10t+(3/2)t^2-t^3-8
not sure if that was correct if anyone could fix my errors!