<PRE><B>Question 976441</B>
a missile is accelerating at a rate of 6t m/sec^2 from a position at rest in a silo 40m above ground level. how high above the ground will it be after 4 seconds
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40 meters above ground?  Most are below ground.
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Accel = 1st derivative of speed = 2nd derivative of displacement.
a = 6t  (assuming t is time in seconds)
Using its starting height as zero at t = 0:
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INT(6t) = 3t^2 + C
C = 0
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INT(3t^2) = t^3 + C
Again, C = 0
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s = t^3
At t = 4, s = 64 meters
Add or subtract the 40 meters
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