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Rudiy27
4 years ago
15

How do you calculate acceleration on a velocity time graph

Physics
2 answers:
adell [148]4 years ago
7 0
Acceleration on a velocity time graph is found by calculating the change in the velocity over the change in time, or in other words, the slope. To do so, it's (final v-initial v)/(final t-initial t), where v is the velocity and t is time. Just find the final point you are looking for on the velocity time graph and determine its velocity and time. Then do the same for the initial point on the velocity time graph. Subtract the final velocity from the initial velocity and divide it by the final time minus the initial time and you will find the velocity for that interval. If you are looking for the instantaneous velocity at one specific point and not a interval, it's just the slope of the line at that point in time.
n200080 [17]4 years ago
5 0

Answer:

\large \boxed{\mathrm{slope \ of \ the \ graph}}

Explanation:

\displaystyle acceleration = \frac{change \ in \ velocity }{elapsed \ time}

\displaystyle acceleration = \frac{\Delta V }{\Delta t}

The average acceleration over a certain time interval tells us by how much the velocity changes per time unit over that interval.

The slope or rise over run of a velocity-time graph tells us about the average acceleration.

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Explanation:

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4 years ago
A _____ is a model of an atom in which each dot represents a valence electron
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Answer:

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Two technicians are discussing a problem where the brake pedal travels too far before the vehicle starts to slow. Technician A s
iVinArrow [24]

Answer:

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3 0
4 years ago
I need an answer for this plzz!!<br>number 2 <br>anybody can help ??
Anuta_ua [19.1K]
2.1) (i) W = mg downwards
(ii) N = R = Normal Reaction from the ground upwards
(iii) Fe = Force of engine towards the right
(iv) f = friction towards the left
(v) ma = Constant acceleration towards right.
2.2.1)
v = 25 m/s
u = 0 m/s
∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s
a \:  =  \:  \frac{dv}{dt}  \:  =  \:  \frac{25 \:  \frac{m}{s} }{50 \: s} \:  =  \: 0.5 \:  \frac{m}{ {s}^{2} }
a = 0.5 m/s².
2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)
2ax \:  =  \:  {v}^{2}  \:  -  \:  {u}^{2}
x \:  =  \:  \frac{ {v}^{2}  \:   -  \:  {u}^{2} }{2a}  \:  =  \:   \frac{{(25 \:  \frac{m}{s})}^{2}  \:  -  \:  {(0 \:  \frac{m}{s} )}^{2} }{2 \:  \times  \: 0.5 \:  \frac{m}{ {s}^{2} } } \:  =  \: 625 \: m
2.3)
Fe = f + ma
Fe - f = ma
For velocity to be constant,
a should be 0, or, a = 0,
Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.
8 0
3 years ago
Describe the friction of smooth surfaces
cluponka [151]

When two surfaces slide against each other, a force called friction makes them stick very slightly together. Smooth surfaces, like ice and glass, are easy to slide over. They create very little friction. Rough surfaces like rock and sand create much more friction, and are easy to grip on to.

hope it helps...!!!

7 0
2 years ago
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