Question #2797

Prove graphically and anatically following equations of motions;

1.) v = u+at

2.)v[sup]2[/sup] = u[sup]2[/sup]+2as

3.)s = ut+1/2at[sup]2[/sup]

1.) v = u+at

2.)v[sup]2[/sup] = u[sup]2[/sup]+2as

3.)s = ut+1/2at[sup]2[/sup]

Expert's answer

Let x(t) be the position of some body. Let v(t) be the velocity of some body. Than its position at any moment of the time will be

∫v_{0} dt = v_{0} t + x_{0} = x(t)

where x_{0} is the initial position.

If body has nonzero acceleration than

at^{2}/2+v_{0} t+x_{0}=x(t)

where a is the acceleration.

If x_{0}=0 we get formula 3.

at^{2}/2 + v_{0} t = x(t) - x_{0} = x(t) = S.

Differentiating this expression we get

x' (t) = v_{0}+at = v(t)

And that is formula 1.

v = v_{0}+at

S = v_{0} t+(at^{2})/2

t = (v-v_{0})/a

S=v_{0} (v-v_{0})/a+(a(v^{2}-2v^{2} v_{0}^{2} + v_{0}^{2}))/(2a^{2} ) = (v^{2}-v_{0}^{2})/2a

a=(v^{2}-v_{0}^{2})/2S

v^{2}=v_{0}^{2}+2Sa.

This is formula 2.

This graph illustrate function

v_{0}t+x_{0}=x(t).

This graph illustrates function

at^{2}/2 + v_{0}t + x_{0} = x(t)

Area under the line for given interval of t gives S.

∫v

where x

If body has nonzero acceleration than

at

where a is the acceleration.

If x

at

Differentiating this expression we get

x' (t) = v

And that is formula 1.

v = v

S = v

t = (v-v

S=v

a=(v

v

This is formula 2.

This graph illustrate function

v

This graph illustrates function

at

Area under the line for given interval of t gives S.

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## Comments

Assignment Expert08.08.11, 13:20You are welcome!

Ayush07.08.11, 12:05I think that your information was much of help to me. Thankx.

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