Question

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

Answer 1

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

Step-by-step 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

$x = \frac{1}{3}t^3 - \frac{1}{2}t^2 - 42t|\limits^{12}_{1}$

Substitute 12 and 1 for t respectively

$x = (\frac{1}{3}*12^3 - \frac{1}{2}*12^2 - 42*12) - (\frac{1}{3}*1^3 - \frac{1}{2}*1^2 - 42*1)$

$x = (576 - 72 - 504) - (\frac{1}{3} - \frac{1}{2} - 42)$

$x = (0) - (-42.17)$

$x = 42.17$