Question

The position coordinates of an object are given by x= 2t2 , y=t2-4t, z=3t-5. The average velocity of the particle during the interval 0.1 sec is?
I think the question is wrong as the answer is given as root22. Can you please rectify the question?????????????

Answer 1

Answer:

The average velocity of the particle is √24 m/s.

Explanation:

Given that,

The position coordinates of an object are

$x=2t^2$ ...(I)

$y=t^2-4t$ ....(II)

$x=3t-5$ .....(III)

Time t = 0.1 sec

The velocity is the rate of change with the time of position of an object.

$v = \dfrac{ds}{dt}$

For x coordinate,

$v_{x}=\dfrac{dx}{dt}=4t$

Put the value of t in the equation

$v_{x} = 0.4\ m/s$

For y coordinate,

$v_{y}=\dfrac{dy}{dt}=2t-4$

Put the value of t in the equation

$v_{y} = 2\times0.1-4=-3.8\ m/s$

For z coordinate,

$v_{z}=\dfrac{dz}{dt}=3$

The average velocity is defined as,

$v_{avg}=\sqrt{v_{x}^+v_{y}^2+v_{z}^2}$

$v_{avg}=\sqrt{(0.4)^2+(3.8)^2+(3)^2}$

$v_{avg}=\sqrt{23.60}\ approximate \sqrt{24}$

Hence,  The average velocity of the particle is √24 m/s.

Answer 2

Answer:

The average velocity of the particle is √24 m/s. ...........................................................................

Explanation:

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

Given that,

The position coordinates of an object are

...(I)

....(II)

.....(III)

Time t = 0.1 sec

The velocity is the rate of change with the time of position of an object.

For x coordinate,

Put the value of t in the equation

For y coordinate,

Put the value of t in the equation

For z coordinate,

The average velocity is defined as,

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Hence,  The average velocity of the particle is √24 m/s.

Explanation: