MHB 2.1.314 AP Calculus Exam a particle moves along the x-axis......

[karush]

View attachment 9347
ok again I used an image since there are macros and image

I know this is a very common problem in calculus but think most still stumble over it
inserted the graph of v(t) and v'(t) and think for v'(t) when the graph is below the x-axis that participle is moving to the left

the integral has a - interval but I think the total is an absolute value...

my take on some of it.

finally this will be my last AP calculus exam question for a while
I was surprized how many views these got... must be a big concern for many

 

[skeeter]

(a) The particle moves toward the left when $v(t) < 0$

$\cos\left(\dfrac{\pi}{6} t \right) < 0 \text{ in the interval } 3 < t < 9$

(b) total distance $$= \int_0^6 |v(t)| \, dt
$$

(c) $a(t) = v'(t) = -\dfrac{\pi}{6} \sin\left(\dfrac{\pi}{6} t \right)$

$a(4) = -\dfrac{\pi}{6} \sin\left(\dfrac{2\pi}{3} \right) < 0$

$v(4) = \cos\left(\dfrac{2\pi}{3} \right) < 0$

Since $a(4)$ and $v(4)$ have the same sign, the particle's speed is increasing

(d) $$x(4) = x(0) + \int_0^4 v(t) \, dt = -2 + \bigg[\dfrac{6}{\pi}\sin\left(\dfrac{\pi}{6} t \right)\bigg]_0^4 = -2+\dfrac{6}{\pi} \bigg[\dfrac{\sqrt{3}}{2} - 0 \bigg]$$