The velocity as a function of time for a car on an amusement park ride is given as v=At^2 + Bt with constants
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The velocity as a function of time for a car on an amusement park ride is given as v=At^2 + Bt with constants

[From: ] [author: ] [Date: 11-09-07] [Hit: ]
what is its position at t=2.0s?-v = t^2 + 2t . Since v = dx/dt , dx/dt = t^2 + 2t .dx = t^2dt + 2tdt .......
I'm terrible with integration, please help!

The velocity as a function of time for a car on an amusement park ride is given as v=At^2 + Bt with constants A=1.0m/s^3 and B=2.0m/s^2. If the car starts at the origin, what is its position at t=2.0s?

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v = t^2 + 2t .
Since v = dx/dt ,
dx/dt = t^2 + 2t .
dx = t^2dt + 2tdt .
Integrating both sides [ integration{ a^nda } = a^(n+1) / (n+1) ]
So , int [ x^0dx ] = int [ t^2dt ] + 2 int [ t^1dt ] ,
Now ,
x^1/1 = t^3/3 + 2*( t^2/2 )
x = t^3/3 + t^2 .
For t = 2 s , x = 8/3 + 4 = 20/3 m .
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