In ΔTUV, v = 240 inches, t = 390 inches and ∠U=71°. Find the area of ΔTUV, to the nearest square inch.

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

Omgggggggggggggggggggggg I need help

Lina20 [59]

Its 5 yd i think. i hope it helped

8 0

2 years ago

Read 2 more answers

Find the volume of the shaded figure by subtracting the smaller volume from the larger

alekssr [168]

Answer:

a. $9a^3 - 9ab^2$

b. $9a(a^2 - b^2)$

Step-by-step explanation:

a.

$Volume = l*w*h$

$Volume_{smaller} = l*w*h$

Where, $l = 9a, w = b, h = b$

$Volume_{smaller} = 9a*b*b = 9ab^2$

$Volume_{larger} = l*w*h$

Where, $l = 9a, w = a, h = a$

$Volume_{smaller} = 9a*a*a = 9a^3$

Volume of the shaded figure = $9a^3 - 9ab^2$

b. $9a^3 - 9ab^2$ expressed in factored form:

Look for the term that is common to 9a³ and 9ab², then take outside the parenthesis.

$9a^3 - 9ab^2 = 9a(a^2 - b^2)$

6 0

3 years ago

An Uber driver charges an initial fee of $15 and $0.50 per mile. James needs a ride to Walmart, which is 6 miles away. How much

sasho [114]

I need points for a test please don’t report

5 0

2 years ago

There are 24 bottles of water in 2 cases of water. If each case of water contains the same number of bottles, what is the minimu

sergiy2304 [10]

Answer:

7

Step-by-step explanation:

yw

5 0

2 years ago

Read 2 more answers