Yes, two integrations are required.
V(t) = integral of a(t) dt from t = 0 to t
V(t) = 1.4 t^2/2 + 3.9t MINUS the value of the same function at t=0 (which is zero), so
V(t) = 1.4 t^2/2 + 3.9t
X(t) - X(0) = integral of V(t) from t = 0 to t
X(0) = 0
X(t) = 1.4 t^3/6 + 1.95 t^2
If t is in seconds, X will be in meters
Practice problem:
A rocket sled for testing equipment under large accelerations starts at rest and accelerates according to the expression:
a= (2.8 m/s^3)t + (3.9 m/s^2)
How far does the rocket move in the time interval t=0 to t=0.81 s?
This is just a practice problem to help me get an equation for my actual problem.
I know acceleration is the 2nd derivative and so I feel like I should integrate this equation (twice?) but my integration abilities are sorely lacking; especially as it regards to including units.
If I integrate how does that affect m/s^3 and m/s^2? Is it a definite integral from 0 to 0.81 s? While I feel I'm on something of the right track here I'm unsure.
Thanks in advance for the help!
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2 answers