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2 years ago

.A particle starts on the origin. It is pushed back to -5.7 m in 2.1 s. Then it is pushed

Physics

2 answers:

harina [27]2 years ago

5 0

Answer: The average velocity is -0.965m/s

Explanation: The first step is to calculate the two velocities is both directions. A velocity is a distance per unit time.

V=d/ t

=-5.7/2.1

=-2.7m/s

For the other direction the velocity is

V=7.3/9.5

=0.77m/s

The average velocity the add the velocities and divide them by 2.

V=-2.7+0.77/2

V= 0.965m/s

GREYUIT [131]2 years ago

3 0

Answer:

The average velocity of the particle is 1.76 m/s.

Explanation:

Given that,

Distance = -5.7 m

Time t = 2.1 s

Forward distance = 7.3 m

Total time = 9.5 sec

We need to calculate the average velocity of the particle

Average velocity :

Average velocity is equal to the displacement divided by change in time.

$v = \dfrac{\Delta D}{\Delta t}$

Where, D = displacement

t = change in time

$v=\dfrac{7.3-(-5.7)}{9.5-2.1}$

$v=1.76\ m/s$

Hence, The average velocity of the particle is 1.76 m/s.

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A car slows down uniformly from a speed of 30.0 m/s to rest in 7.20 s

Ede4ka [16]

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where $v$ and $v_0$ are the final and initial velocities, respectively. By definition, we also have that the average velocity is given by

$\bar v=\dfrac{\Delta x}{\Delta t}=\dfrac{x-x_0}{t-t_0}$

where $x,x_0$ are the final/initial displacements, and $t,t_0$ are the final/initial times, respectively.

Take the car's starting position to be at $t_0=0\,\mathrm s$ . Then

$\dfrac{v+v_0}2=\dfrac{x-x_0}t\implies x=x_0+\dfrac12(v+v_0)t$

So we have

$x=0\,\mathrm m+\dfrac12\left(0\,\dfrac{\mathrm m}{\mathrm s}+30.0\,\dfrac{\mathrm m}{\mathrm s}\right)(7.20\,\mathrm s)=108\,\mathrm m$

You also could have first found the acceleration using the equation

$v=v_0+at$

then solve for $x$ via

$x=x_0+v_0t+\dfrac12at^2$

but that would have involved a bit more work, and it turns out we didn't need to know the precise value of $a$ anyway.

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Answer:

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Given

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