An object has a mass of $4 k g$ $4 kg$ . The object's kinetic energy uniformly changes from $640 K J$ $640 KJ$ to $160 K J$ $160 KJ$ over $t \in [0, 12 s]$ $t in [0, 12 s]$ . What is the average speed of the object?

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An object has a mass of $4 k g$ $4 kg$ . The object's kinetic energy uniformly changes from $640 K J$ $640 KJ$ to $160 K J$ $160 KJ$ over $t \in [0, 12 s]$ $t in [0, 12 s]$ . What is the average speed of the object?

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Answer 1 · 2017-04-11T16:34:34

The average speed is $= 440 m s^{- 1}$ $=440ms^-1$

Explanation:

The kinetic energy is

$K E = \frac{1}{2} m v^{2}$ $KE=1/2mv^2$

mass is $= 4 k g$ $=4kg$

The initial velocity is $= u_{1}$ $=u_1$

$\frac{1}{2} m u_{1}^{2} = 640000 J$ $1/2m u_1^2=640000J$

The final velocity is $= u_{2}$ $=u_2$

$\frac{1}{2} m u_{2}^{2} = 160000 J$ $1/2m u_2^2=160000J$

Therefore,

$u_{1}^{2} = \frac{2}{4} \cdot 640000 = 320000 m^{2} s^{- 2}$ $u_1^2=2/4*640000=320000m^2s^-2$

and,

$u_{2}^{2} = \frac{2}{4} \cdot 160000 = 80000 m^{2} s^{- 2}$ $u_2^2=2/4*160000=80000m^2s^-2$

The graph of $v^{2} = f (t)$ $v^2=f(t)$ is a straight line

The points are $(0, 320000)$ $(0,320000)$ and $(12, 80000)$ $(12,80000)$

The equation of the line is

$v^{2} - 320000 = \frac{80000 - 320000}{12} t$ $v^2-320000=(80000-320000)/12t$

$v^{2} = - 20000 t + 320000$ $v^2=-20000t+320000$

So,

$v = \sqrt{(- 20000 t + 320000)}$ $v=sqrt((-20000t+320000)$

We need to calculate the average value of $v$ $v$ over $t \in [0, 12]$ $t in [0,12]$

$(12 - 0) ¯ v = \int_{0}^{12} \sqrt{(- 20000 t + 320000)} d t$ $(12-0)bar v=int_0^12sqrt((-20000t+320000))dt$

$12 ¯ v = ⎡ ⎢ ⎣ {⎛ ⎝ \frac{{(- 20000 t + 320000)}^{\frac{3}{2}}}{- \frac{3}{2} \cdot 20000} ⎤ ⎦}_{0}^{12}$ $12 barv=[((-20000t+320000)^(3/2)/(-3/2*20000)]_0^12$

$= ⎛ ⎝ \frac{{(- 20000 \cdot 12 + 320000)}^{\frac{3}{2}}}{- 30000} ⎞ ⎠ - ⎛ ⎝ \frac{{(- 20000 \cdot 0 + 320000)}^{\frac{3}{2}}}{- 30000} ⎞ ⎠$ $=((-20000*12+320000)^(3/2)/(-30000))-((-20000*0+320000)^(3/2)/(-30000))$

$= \frac{320000^{\frac{3}{2}}}{30000} - \frac{80000^{\frac{3}{2}}}{30000}$ $=320000^(3/2)/30000-80000^(3/2)/30000$

$= 5279.7$ $=5279.7$

So,

$¯ v = \frac{5279.7}{12} = 440 m s^{- 1}$ $barv=5279.7/12=440ms^-1$

Answer 2 · 2017-04-11T16:36:31

average speed is defined as total distance travelled divided by total time taken to cover that distance .

Explanation:

the kinetic energy of a body of mass m moving with speed v is defined as : $\frac{1}{2} m v^{2}$ $1/2mv^2$
thus you can calculate velocity of body at the starting and the end of interval of time :
lets say initially the velocity was $v_{1}$ $v_1$ , $\frac{1}{2} \cdot 4 \cdot v_{1}^{2} = 640000$ $1/2*4*v_1^2 = 640000$
$v_{1} = 565.68 m s^{- 1}$ $v_1 = 565.68 ms^-1$
let at the end of interval of time the velocity was $v_{2}$ $v_2$ , $\frac{1}{2} \cdot 4 \cdot v_{2}^{2} = 160000$ $1/2 * 4 * v_2^2 = 160000$
$v_{2} = 282.84 m s^{- 1}$ $v_2 = 282.84 ms^-1$
now , total distance travelled by the body over given time interval can be calculated by using $v_{2}^{2} - v_{1}^{2} = 2 \cdot a \cdot s$ $v_2^2 - v_1^2 = 2*a*s$ and $v_{2} = v_{1} + a \cdot t$ $v_2 = v_1 + a*t$
thus, $282.84 = 565.68 + 12 a$ $282.84 = 565.68 + 12a$
$a = - 23.57 m s^{- 2}$ $a= -23.57ms^-2$
${282.84}^{2} - {565.68}^{2} = 2 \cdot - 23.57 \cdot s$ $282.84^2 - 565.68^2 = 2*-23.57*s$
$s = 5091.12 m$ $s= 5091.12m$
average speed is defined as ratio of total distance travelled and total time taken :
$v = \frac{s}{t}$ $v= s/t$
$v = \frac{5091.12}{12}$ $v= 5091.12/12$
$v = 424.26 m s^{- 1}$ $v= 424.26ms^-1$

Answer 3 · 2017-04-12T04:19:50

Let rate of change of kinetic energy be
$\frac{d K E (t)}{d t} = C$ $(dKE(t))/(dt) = C$
where $C$ $C$ is constant with time $t$ $t$ .

Integrating both sides with $t$ $t$ we get
$\int \frac{d K E (t)}{d t} \cdot d t = \int C \cdot d t$ $int (dKE(t))/(dt) cdot dt = int C cdotdt$

$\Rightarrow K E = C \cdot t + c$ $=>KE= Ccdott + c$ ......(1)
where $c$ $c$ is a constant of integration.

To evaluate $c$ $c$ , use initial condition at $t = 0$ $t=0$
$640 \times 10^{3} = C \cdot 0 + c$ $640xx10^3 = Ccdot0+c$
$\Rightarrow c = 640 \times 10^{3}$ $=>c=640xx10^3$
We have expression for kinetic energy as
$K E = C \cdot t + 640 \times 10^{3}$ $KE= Ccdott + 640xx10^3$

To evaluate $C$ $C$ , use final condition at $t = 12$ $t=12$
$160 \times 10^{3} = C \cdot 12 + 640 \times 10^{3}$ $160 xx10^3= Ccdot12+640xx10^3$
$\Rightarrow C = - 40 \times 10^{3}$ $=>C=-40xx10^3$

Equation (1) becomes
$K E = (- t + 16) \times 4 \times 10^{4}$ $KE= (-t + 16)xx4xx10^4$ ......(2)

If $m$ $m$ is the mass and $v (t)$ $v(t)$ is the velocity of body, Kinetic energy is given as
$K E = \frac{1}{2} m v^{2}$ $KE=1/2mv^2$
$\Rightarrow \frac{1}{2} m v^{2} (t) = (- t + 16) \times 4 \times 10^{4}$ $=> 1/2mv^2(t) = (-t + 16)xx4xx10^4$
Given is $m = 4 k g$ $m=4kg$ . Above expression becomes
$\Rightarrow \frac{1}{2} \times 4 v^{2} (t) = (- t + 16) \times 4 \times 10^{4}$ $=> 1/2xx4v^2(t) = (-t + 16)xx4xx10^4$
$\Rightarrow v {(t)}^{2} = (- t + 16) \times 2 \times 10^{4}$ $=> v(t)^2 = (-t+16)xx2xx10^4$
$\Rightarrow v = \pm \sqrt{(- t + 16) \times 2 \times 10^{4}}$ $=> v= +-sqrt((-t+16)xx2xx10^4)$
$\Rightarrow speed | v | = \sqrt{(- t + 16) \times 2 \times 10^{4}}$ $=> "speed "|v|= sqrt((-t+16)xx2xx10^4)$

Average speed $= \frac{Total Distance traveled}{Time taken} = \frac{\int_{0}^{12} | v (t) | \cdot d t}{12 - 0}$ $="Total Distance traveled"/"Time taken"=(int_0^12|v(t)|*dt)/(12-0)$

$:.$ Average speed $= (int_(0)^(12)sqrt((-t+16)xx2xx10^4)*dt)/12$
$= 100sqrt(2)/12[-(16-t)^(3/2)/(3/2)]_0^12$
$=- 50/9sqrt(2)(4^(3/2) - 16^(3/2))$
$= 439.98 ms^-1$ , rounded to two decimal places.

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An object has a mass of $4 k g$ $4 kg$ . The object's kinetic energy uniformly changes from $640 K J$ $640 KJ$ to $160 K J$ $160 KJ$ over $t \in [0, 12 s]$ $t in [0, 12 s]$ . What is the average speed of the object?

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An object has a mass of $4 k g$ $4 kg$ . The object's kinetic energy uniformly changes from $640 K J$ $640 KJ$ to $160 K J$ $160 KJ$ over $t \in [0, 12 s]$ $t in [0, 12 s]$ . What is the average speed of the object?