Direction: Match the exponential notation in Column A to its

Note the use of a closed or filled in circle at the endpoint (in contrast to an open circle). This indicates we are including the endpoint 19 as part of the solution set, and that's why we go for "or equal to" as part of the inequality sign.

5 0

2 years ago

NO LINKS OR FILES!

Archy [21]

(a) If the particle's position (measured with some unit) at time t is given by s(t), where

$s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}$

then the velocity at time t, v(t), is given by the derivative of s(t),

$v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}$

(b) The velocity after 3 seconds is

$v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}$

(c) The particle is at rest when its velocity is zero:

$\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}$

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

$\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|$

In interval notation, this happens for t in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt$

By definition of absolute value, we have

$|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)$

In part (d), we've shown that v(t) > 0 when -√11 < t < √11, so we split up the integral at t = √11 as

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt$

and by the fundamental theorem of calculus, since we know v(t) is the derivative of s(t), this reduces to

$s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}$

7 0

2 years ago