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# Suppose that an object is at position s(t)=t^2 feet at time t seconds.

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Suppose that an object is at position s(t)=t^2 feet at time t seconds.

A.)Find the average velocity of the object over a time interval from time t seconds to time 2 seconds.

B.)Find the instantaneous velocity of the object at time 2 seconds by taking the limit of the average velocity in part A as t--> 2.

Guest Jan 30, 2016

#2
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Now,I want to show you guys some physics and mathematics.

acceleration is the change in velocity in a given amount of time.$$a=\frac{V-Vo}{t}$$

where V is the final veocity and Vo is the initial velocity.

rearrange,we have $$V=at+Vo$$If the object's accelration is constant,then the graph of the velocity is a linear function.

But I would like to express it in y=mx+b form in my graph.And here is my graph

The area of the triangle is the sum of the  two base times the hight divide by two.or A=(b1+b2)*h/2

In here ,we could express it as $$A=(V+Vo)*\Delta t/2$$

the area under the curve(line) of function of velocity is the displacement.

Therefore,$$\Delta S=(V+Vo)*\Delta t/2$$$$a=\frac{V-Vo}{t}\Rightarrow t=\frac{V-Vo}{a}$$

If the object start moving at time 0 seconds

$$\Delta t=T-To=t$$

the displacement of the object from time 0 seonds is $$\Delta S=\frac{V^2-Vo^2}{2a}$$

substitute V=at+Vo into $$\Delta S=(V+Vo)*\Delta t/2$$

we have $$\Delta S=(2Vo+at)*t/2\Rightarrow \Delta S=1/2at^2+Vo*t$$

displacement equal final position subract inital position $$\Delta S=S-So$$

$$S=\Delta S+So$$$$S=1/2at^2+Vo*t+So$$

Confirm my previous equation by using integration

$$V=\frac{ds}{dt}=a*t+Vo \Rightarrow \int \frac{ds}{dt}dt=\int at+Vo dt\Rightarrow S=1/2at^2+Vo*t+C$$ for some costant C

In here C is the initial position,so $$S=1/2at^2+Vo*t+So$$

http://www.desmos.com/calculator/ibbxgibq5n (my original graph from demos)

fiora  Jan 31, 2016
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#1
+519
+10

The instantaneous velocity is  the derivative of the position function s=f(t) with respect to time.At tiem t,the velocity is

V(t)=$$\frac{ds}{dt}=\lim_{x\rightarrow \Delta t } \frac{S(t+\Delta t)-S(t)}{\Delta t}$$

(1)v(t)=$$\frac{ds}{dt} t^2=2t$$

(2)$$\lim_{t\rightarrow 2} 2t=2*2=4$$

not good at calculus,might be wrong

fiora  Jan 30, 2016
#2
+519
+15

Now,I want to show you guys some physics and mathematics.

acceleration is the change in velocity in a given amount of time.$$a=\frac{V-Vo}{t}$$

where V is the final veocity and Vo is the initial velocity.

rearrange,we have $$V=at+Vo$$If the object's accelration is constant,then the graph of the velocity is a linear function.

But I would like to express it in y=mx+b form in my graph.And here is my graph

The area of the triangle is the sum of the  two base times the hight divide by two.or A=(b1+b2)*h/2

In here ,we could express it as $$A=(V+Vo)*\Delta t/2$$

the area under the curve(line) of function of velocity is the displacement.

Therefore,$$\Delta S=(V+Vo)*\Delta t/2$$$$a=\frac{V-Vo}{t}\Rightarrow t=\frac{V-Vo}{a}$$

If the object start moving at time 0 seconds

$$\Delta t=T-To=t$$

the displacement of the object from time 0 seonds is $$\Delta S=\frac{V^2-Vo^2}{2a}$$

substitute V=at+Vo into $$\Delta S=(V+Vo)*\Delta t/2$$

we have $$\Delta S=(2Vo+at)*t/2\Rightarrow \Delta S=1/2at^2+Vo*t$$

displacement equal final position subract inital position $$\Delta S=S-So$$

$$S=\Delta S+So$$$$S=1/2at^2+Vo*t+So$$

Confirm my previous equation by using integration

$$V=\frac{ds}{dt}=a*t+Vo \Rightarrow \int \frac{ds}{dt}dt=\int at+Vo dt\Rightarrow S=1/2at^2+Vo*t+C$$ for some costant C

In here C is the initial position,so $$S=1/2at^2+Vo*t+So$$

http://www.desmos.com/calculator/ibbxgibq5n (my original graph from demos)

fiora  Jan 31, 2016
#3
+519
+5
fiora  Jan 31, 2016
edited by fiora  Jan 31, 2016
edited by fiora  Jan 31, 2016
#4
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I think you should give yourself some points !