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The velocity function, in feet per second, is given for a particle moving along a straight line. () = 2 − − 42, 1 ≤ ≤ 12

User Archaelus
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Question:

The velocity function, in feet per second, is given for a particle moving along a straight line.
v(t) = t^2 - t - 42
1 \le t \le 12

Find the displacement

Answer:

The displacement is 42.17ft

Explanation:

Given


v(t) = t^2 - t - 42
1 \le t \le 12

The displacement x, is calculated using:


x = \int\limits^a_b {v(t)} \, dt


x = \int\limits^(12)_(1) {t^2 - t - 42} \, dt

Integrate


x = (1)/(3)t^3 - (1)/(2)t^2 - 42t|\limits^(12)_(1)

Substitute 12 and 1 for t respectively


x = ((1)/(3)*12^3 - (1)/(2)*12^2 - 42*12) - ((1)/(3)*1^3 - (1)/(2)*1^2 - 42*1)


x = (576 - 72 - 504) - ((1)/(3) - (1)/(2) - 42)


x = (0) - (-42.17)


x = 42.17

User Daniel Cukier
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