Please help with Calculus

Question

Find the distance, in feet, a particle travels in its first 2 seconds of travel, if it moves according to the velocity equation v(t)= 6t² - 18t + 12 (in feet/sec)

Options:

1. 4
2. 5
3. 6
4. -1

Answer 1

Hello Mikaela! I'll help you out!

Remember that distance = integral(t=a,t=b, | v(t) |, dt)

Note that the v(t) is positive for 0 < t < 1, and v(t) is negative for 1 < t < 2. We can break this down to:

integral(0, 1, v(t), dt) + ( -1 * integral(1, 2, v(t), dt) ) >>> Note how the negative of a negative area is positive, and so it replaces the absolute value.

Using antiderivatives, you should find that integral(0, 1, v(t), dt) = 5 and that integral(1, 2, v(t), dt) = -1, so:

5 - -1 = 6 feet. Therefore, (3) <<<<<<<<<< (ANSWER). Please let me know if you need additional help!