An object has a mass of $6 k g$ $6 kg$ . The object's kinetic energy uniformly changes from $540 K J$ $540 KJ$ to $32 K J$ $32 KJ$ over $t \in [0, 4 s]$ $t in [0, 4 s]$ . What is the average speed of the object?

Question

An object has a mass of $6 k g$ $6 kg$ . The object's kinetic energy uniformly changes from $540 K J$ $540 KJ$ to $32 K J$ $32 KJ$ over $t \in [0, 4 s]$ $t in [0, 4 s]$ . What is the average speed of the object?

1 Answer

Impact of this question

1563 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-24T17:15:12

The average speed is $= 296.3 m s^{- 1}$ $=296.3ms^-1$

Explanation:

The kinetic energy is

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

The mass is $= 6 k g$ $=6kg$

The initial velocity is $= u_{1} m s^{- 1}$ $=u_1ms^-1$

The final velocity is $= u_{2} m s^{- 1}$ $=u_2 ms^-1$

The initial kinetic energy is $\frac{1}{2} m u_{1}^{2} = 540000 J$ $1/2m u_1^2=540000J$

The final kinetic energy is $\frac{1}{2} m u_{2}^{2} = 32000 J$ $1/2m u_2^2=32000J$

Therefore,

$u_{1}^{2} = \frac{2}{6} \cdot 540000 = 180000 m^{2} s^{- 2}$ $u_1^2=2/6*540000=180000m^2s^-2$

and,

$u_{2}^{2} = \frac{2}{6} \cdot 32000 = 10666.7 m^{2} s^{- 2}$ $u_2^2=2/6*32000=10666.7m^2s^-2$

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

The points are $(0, 180000)$ $(0,180000)$ and $(4, 10666.7)$ $(4,10666.7)$

The equation of the line is

$v^{2} - 180000 = \frac{10666.7 - 180000}{4} t$ $v^2-180000=(10666.7-180000)/4t$

$v^{2} = - 42333.3 t + 180000$ $v^2=-42333.3t+180000$

So,

$v = \sqrt{(- 42333.3 t + 180000)}$ $v=sqrt((-42333.3t+180000)$

We need to calculate the average value of $v$ $v$ over $t \in [0, 4]$ $t in [0,4]$

$(4 - 0) ¯ v = \int_{0}^{4} (\sqrt{- 42333.3 t + 180000}) d t$ $(4-0)bar v=int_0^4(sqrt(-42333.3t+180000))dt$

$4 ¯ v = {⎡ ⎣ ⎛ ⎝ \frac{{(- 42333.3 t + 180000)}^{\frac{3}{2}}}{- \frac{3}{2} \cdot 42333.3} ⎞ ⎠ ⎤ ⎦}_{0}^{4}$ $4 barv=[((-42333.3t+180000)^(3/2)/(-3/2*42333.3))]_0^4$

$= ⎛ ⎝ \frac{{(- 42333.3 \cdot 4 + 180000)}^{\frac{3}{2}}}{- 63500} ⎞ ⎠ - ⎛ ⎝ \frac{{(42333.3 \cdot 0 + 180000)}^{\frac{3}{2}}}{- 63500} ⎞ ⎠$ $=((-42333.3*4+180000)^(3/2)/(-63500))-((42333.3*0+180000)^(3/2)/(-63500))$

$= \frac{180000^{\frac{3}{2}}}{63500} - \frac{{10666.7}^{\frac{3}{2}}}{63500}$ $=180000^(3/2)/63500-10666.7^(3/2)/63500$

$= 1185.3$ $=1185.3$

So,

$¯ v = \frac{1185.3}{4} = 296.3 m s^{- 1}$ $barv=1185.3/4=296.3ms^-1$

The average speed is $= 296.3 m s^{- 1}$ $=296.3ms^-1$

Science

Math

Humanities

... and beyond

An object has a mass of $6 k g$ $6 kg$ . The object's kinetic energy uniformly changes from $540 K J$ $540 KJ$ to $32 K J$ $32 KJ$ over $t \in [0, 4 s]$ $t in [0, 4 s]$ . What is the average speed of the object?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

An object has a mass of 6 kg6kg6 kg. The object's kinetic energy uniformly changes from 540 KJ540KJ540 KJ to 32 KJ32KJ 32 KJ over t in [0, 4 s]t∈[0,4s]t in [0, 4 s]. What is the average speed of the object?

1 Answer

Explanation:

Related questions

Impact of this question

An object has a mass of $6 k g$ $6 kg$ . The object's kinetic energy uniformly changes from $540 K J$ $540 KJ$ to $32 K J$ $32 KJ$ over $t \in [0, 4 s]$ $t in [0, 4 s]$ . What is the average speed of the object?