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# Kinematic equations derived from Calculus.

First important relation:

Consider the graph of a function $$\displaystyle f(t)$$ 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 $$\displaystyle t_i$$ and $$\displaystyle t_f$$ is: $\Delta_x=x_f-x_i=\int_{t_i}^{t_f}f(t)\,\mathrm dt.$ The same thing can be performed with a function $$\displaystyle g(t)$$ that outputs for each time, the acceleration at that time. And so the area under the graph of this function between time $$\displaystyle t_i$$ and $$\displaystyle t_f$$ is: $\Delta_{v}=v_{xf}-v_{xi}=\int_{t_i}^{t_f}g(t)\,\mathrm dt.$

Kinematics!

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

In the same spirit, we can derive another kinematics equation by recalling that velocity is: $v_{x}=\dfrac{\mathrm dx}{\mathrm dt}.$ Rearranging we get $$\displaystyle \mathrm dx=v_{x}\,\mathrm dt.$$ If we integrate both sides and using our very first relation we get that: $\Delta_x=x_f-x_i=\int_{0}^t v_{x}\,\mathrm dt.$ And since $$\displaystyle v_{x}=v_{xf}=v_{xi}+a_xt$$ 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 $$\displaystyle x_i$$ to both sides we get: $\color{grey}{\boxed{\color{white}{\overline{\color{white}{\underline{\color{black}{ x_{f}=x_i+v_{xi}t+\tfrac12a_x\,t^2. }}}}}}}$ This relation let us determine the final position $$\displaystyle x_f$$ at time $$\displaystyle t$$ if we know the initial position $$\displaystyle x_i$$ with the initial velocity $$\displaystyle x_i$$ and if the acceleration $$\displaystyle a_x$$ is constant and well-known.

I hope this helps. Best wishes, $$\displaystyle \cal H$$akim.

Note by حكيم الفيلسوف الضائع
2 years, 9 months ago

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شكرا ولد لبلاد · 2 years, 6 months ago

hhh :-D · 2 years, 6 months ago

Wow excellent thanks I didn't know this · 2 years, 9 months ago

Glad you liked it! $$\overset{\cdot\cdot}\smile$$ · 2 years, 9 months ago

Nice explanation · 2 years, 9 months ago

Glad you liked it! $$\overset{\cdot\cdot}\smile$$ · 2 years, 9 months ago

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. · 2 years, 8 months ago

this is good...basic of kinematics ..thanks for recalling.. · 2 years, 8 months ago

We can even derive Torricelli's Equations, such as $$v_f^{2}-v_i^{2}=2as$$. · 2 years, 9 months ago

Yea I agree by the way what grade are you in · 2 years, 9 months ago

11th · 2 years, 9 months ago