going from a(t) to v(t): Where did I go wrong?

insanerp

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The acceleration function a(t) in m/s^2, a(t)=2t + 4, 0<equal to(t) <equal to 3
initial velocity v(0), both are given for a particle moving along a line v(0)=-5

find v(t) at time t, and find the distance d traveled during the time interval?

So I took the integral of a(t)which is = t^2 +4t to find v(t) and 18m/s and the distance would be 18 X 3sec. -v(0)=54-5=49m

Where did I go wrong?
Thanks,
Michele
 
insanerp said:
The acceleration function a(t) in m/s^2, a(t)=2t + 4.... v(0)=-5....

I took the integral of a(t)which is = t^2 +4t
Your work is a bit confusing, but at a guess: Did you forget the constant of integration...? Did you forget to use the initial value, v(0) = -5, to find the value of the constant of integration...? :oops:

Eliz.
 
insanerp said:
The acceleration function a(t) in m/s^2, a(t)=2t + 4, 0<equal to(t) <equal to 3
initial velocity v(0), both are given for a particle moving along a line v(0)=-5

find v(t) at time t, and find the distance d traveled during the time interval?

So I took the integral of a(t)which is = t^2 +4t to find v(t) and 18m/s and the distance would be 18 X 3sec. -v(0)=54-5=49m

Where did I go wrong?
Thanks,
Michele

The equation for speed should be

v(t) = t^2 + 4t - 5

You need to integrate this again to get d(t).

d = v * t won't work here because v(t) is not a constant.
 
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