Kinematic equations derived from Calculus.

First important relation:

Consider the graph of a function f(t)\displaystyle f(t) which takes as an input time and outputs velocity at that particular time. From calculus we know that the area under the graph of this function between time ti\displaystyle t_i and tf\displaystyle t_f is: Δx=xfxi=titff(t)dt.\Delta_x=x_f-x_i=\int_{t_i}^{t_f}f(t)\,\mathrm dt. The same thing can be performed with a function g(t)\displaystyle g(t) that outputs for each time, the acceleration at that time. And so the area under the graph of this function between time ti\displaystyle t_i and tf\displaystyle t_f is: Δv=vxfvxi=titfg(t)dt.\Delta_{v}=v_{xf}-v_{xi}=\int_{t_i}^{t_f}g(t)\,\mathrm dt.

Kinematics!

Acceleration is defined as ax=dvxdt.a_x=\dfrac{\mathrm d v_x}{\mathrm d t}. We can rearrange it to get dvx=axdt.\displaystyle \mathrm dv_x=a_x\,\mathrm dt. If we integrate both sides and consider our first relation, and letting ti=0\displaystyle t_i=0 and tf=t\displaystyle t_f=t, we get that: vxfvxi=0taxdt.v_{xf}-v_{xi}=\int_0^t a_x\,\mathrm dt. And if acceleration is constant, and using the fundamental theorem of calculus, we get that: vxfvxi=0taxdt=ax(t0)=axt.v_{xf}-v_{xi}=\int_0^t a_x\,\mathrm dt=a_x(t-0)=a_xt. If we add vxi\displaystyle v_{xi} to both sides we obtain: vxf=vxi+axt.\color{grey}{\boxed{\color{#FFFFFF}{\overline{\color{#FFFFFF}{\underline{\color{#333333}{ v_{xf}=v_{xi}+a_x\,t. }}}}}}} This relation will let us find the final velocity at time t\displaystyle t if we know the initial velocity vxi\displaystyle v_{xi} and the (constant) acceleration ax\displaystyle a_x.

In the same spirit, we can derive another kinematics equation by recalling that velocity is: vx=dxdt.v_{x}=\dfrac{\mathrm dx}{\mathrm dt}. Rearranging we get dx=vxdt.\displaystyle \mathrm dx=v_{x}\,\mathrm dt. If we integrate both sides and using our very first relation we get that: Δx=xfxi=0tvxdt.\Delta_x=x_f-x_i=\int_{0}^t v_{x}\,\mathrm dt. And since vx=vxf=vxi+axt\displaystyle v_{x}=v_{xf}=v_{xi}+a_xt we can rewrite our definite integral as \eqalign{ x_f-x_i&=\int_{0}^t v_{x}\,\mathrm dt \\ &=\int_0^t\left(v_{xi}+a_xt\right)\mathrm dt \\ &= \int_0^tv_{xi}\,\mathrm dt+a_x\int_0^t t\,\mathrm dt \\ &=v_{xi}t+a_x\dfrac{t^2}{2} \\ &=v_{xi}+\tfrac12a_xt^2.} Adding xi\displaystyle x_i to both sides we get: xf=xi+vxit+12axt2.\color{grey}{\boxed{\color{#FFFFFF}{\overline{\color{#FFFFFF}{\underline{\color{#333333}{ x_{f}=x_i+v_{xi}t+\tfrac12a_x\,t^2. }}}}}}} This relation let us determine the final position xf\displaystyle x_f at time t\displaystyle t if we know the initial position xi\displaystyle x_i with the initial velocity xi\displaystyle x_i and if the acceleration ax\displaystyle a_x is constant and well-known.

I hope this helps. Best wishes, \calH\displaystyle \cal Hakim.

Note by حكيم الفيلسوف الضائع
5 years, 4 months ago

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شكرا ولد لبلاد

مغربي بسيط - 5 years ago

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Nice explanation

Laith Hameed - 5 years, 4 months ago

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Glad you liked it! \overset{\cdot\cdot}\smile

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Wow excellent thanks I didn't know this

Mardokay Mosazghi - 5 years, 4 months ago

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Glad you liked it! \overset{\cdot\cdot}\smile

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What about the equation

Vf^2=vo^2+2as? Put that in your discussion.

Bonus: try incorporating the chain rule.

Kao Cen Darach - 11 months, 1 week ago

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We can even derive Torricelli's Equations, such as vf2vi2=2asv_f^{2}-v_i^{2}=2as.

Nanayaranaraknas Vahdam - 5 years, 4 months ago

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Yea I agree by the way what grade are you in

Mardokay Mosazghi - 5 years, 4 months ago

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11th

Nanayaranaraknas Vahdam - 5 years, 4 months ago

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this is good...basic of kinematics ..thanks for recalling..

Akash Kumar - 5 years, 3 months ago

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equation of motion already explained in INDIAN NCERT book by this method but not in perfect way .here this one is explained in perfect way.

Kundan Kumar - 5 years, 3 months ago

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No meaning of level 3. These derivations are known to every 9th standard student.

Anshul Yadav - 5 years, 2 months ago

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