# The acceleration of a point moving in a straight line is given by the equation a = 6t + 12.

The acceleration of a point moving in a straight line is given by the equation a = 6t + 12. Find the path traversed by the point in 3 s, if at the moment of time t = 2c the point has the speed U = 38 m / s and has passed the path s = 30 m.

Knowing the law of variation of acceleration with respect to time, we find the law of variation of speed with respect to time, i.e. we find the antiderivative of acceleration in time:

v (t) = 3 * t² + 12 * t + C.

Because v (2) = 38, then 3 * 2² + 12 * 2 + C = 38,

C = 38 – 12 – 24 = 2.

Hence, the instantaneous velocity function:

v (t) = 3 * t² + 12 * t + 2.

The distance traveled is the integral of the speed over time, therefore, we get:

s = integral (from 0 to 3) (3 * t² + 12 * t + 2) dt = t³ + 6 * t² + 2 * t (from 0 to 3) = 27 + 54 + 6 = 87 m.

Answer: the material point has passed 87 m.

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