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1. An 2 kg object has an inital velocity of 4 m/s. If it comes to a stop in a distance of 12 meters.

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1. An 2 kg object has an inital velocity of 4 m/s. If it comes to a stop in a distance of 12 meters.

What was the acceleration of the object.

time is 12 / 4 = 3

12 / 3 = 4

would 4 be the acceleration?

2. If you throw a ball at 30 degrees into the air with a velocity of 18 m/s. How long is it in the air?

3. A 4kg object is 30m above the ground.

If it falls what is the object velocity 2 meters above the ground?

Guest Feb 26, 2017
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8+0 Answers

#1
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For the first one, I'm not really sure but I think that this is how to do it.

Imagine a graph of the velocity of the object.

Its velocity is 4 m/s at x = 0, and its velocity is 0 m/s at x=12 meters.

So if y is the velocity and x is the distance, the equation of that line is: y = (-4/12)x + 4, or

y = (-1/3)x + 4.

The derivative of velocity is acceleration, so the acceleration is -(1/3) m/s/s.

You know the acceleration is going to be negative because the object is slowing down.

I'm not sure if this is right though because I didn't use the information about it being 2 kg at all.

hectictar  Feb 26, 2017
#2
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I also want to attempt the second one because why not haha I have no clue if I'm doing these right though.

So I'm going to work towards getting a position function of the ball.

Acceleration:

y''= -9.8 m/s2

Velocity:

y'=$$\int -9.8 dx$$

y' = -9.8x + c

18sin30° = -9.8(0) + c

c=9

y'= -9.8x + 9

Position:

y=$$\int(-9.8x+9)dx$$

y=$$\int-9.8x dx +\int9dx$$

y=$$-9.8\frac{x^{2}}{2}+9x$$

y= -4.9x2 + 9x

So that is the position function of the ball. So we need to find the two places where the ball touches the ground, aka where the y value equals 0.

0 = -4.9x2 + 9x

0 = x(-4.9x+9)

Set each factor equal to zero.

-4.9x+9 = 0 and x = 0

-4.9x=-9

x = 90/49

So y is 0 when x is 0 and when x is 90/49. The difference in those x values will be the distance it  traveled. So it traveled 90/49 meters. I just realized that we want the time it is in the air, not the distance traveled. Well it's okay the time it was in the air will just be 18 m/s divided by 90/49 meters. Which is 49/5 or 9.8 seconds. That was probably the longest possible approach to this problem but I kind of think I got the right answer.

hectictar  Feb 26, 2017
#5
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Oh okay mine is wrong because I found the displacement not the distance, so when I tried to convert that to air time it just gave me something that was meaningless. I didn't really know what I was doing, heh.

But I'm confused though because when I tried to find the arc length from 0 to 90/49 I got about 8.61142 and then 18 divided by that is about 2.09. Howcome 2.09 seconds isn't the answer?

hectictar  Feb 26, 2017
#3
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2. If you throw a ball at 30 degrees into the air with a velocity of 18 m/s. How long is it in the air?

We are using :

t  = 2 v0 sin θ / g

t =  2* 18m/s * sin (30)  / [9/8m /s^2]  =

36 * (1/2)  / 9.8    ≈  1.837 s

Here's a graph : https://www.desmos.com/calculator/2i436tqxnr

CPhill  Feb 26, 2017
#4
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3. A 4kg object is 30m above the ground.

If it falls what is the object velocity 2 meters above the ground?

The time it takes to fall 28m  is :

-28  = -(1/2)9.81 * t^2

56/9.8  = t^2

sqrt (56/9.8)  = t ≈  2.39 s

And the velocity  is given by :

v(2.39)  =  9.8m/s^2 (2.39 s)   ≈ 23.4 m/s

CPhill  Feb 26, 2017
#6
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I don't know, hectictar.....I'm not really a "Physics" person....

BTW....to be technically correct.....the answer to (3)  should be - 23.4 m/s

CPhill  Feb 26, 2017
#7
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Oh okay, thanks

hectictar  Feb 26, 2017
edited by hectictar  Oct 17, 2017
#8
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Constant acceleration kinetic equations can be used, as follows:

.

Alan  Feb 26, 2017

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