**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.

## Comments

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TopNewestشكرا ولد لبلاد – مغربي بسيط · 2 years, 6 months ago

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– حكيم الفيلسوف الضائع · 2 years, 6 months ago

hhh :-DLog in to reply

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

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– حكيم الفيلسوف الضائع · 2 years, 9 months ago

Glad you liked it! \(\overset{\cdot\cdot}\smile\)Log in to reply

Nice explanation – Laith Hameed · 2 years, 9 months ago

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– حكيم الفيلسوف الضائع · 2 years, 9 months ago

Glad you liked it! \(\overset{\cdot\cdot}\smile\)Log in to reply

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 · 2 years, 8 months ago

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

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We can even derive Torricelli's Equations, such as \(v_f^{2}-v_i^{2}=2as\). – Nanayaranaraknas Vahdam · 2 years, 9 months ago

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– Mardokay Mosazghi · 2 years, 9 months ago

Yea I agree by the way what grade are you inLog in to reply

– Nanayaranaraknas Vahdam · 2 years, 9 months ago

11thLog in to reply

No meaning of level 3. These derivations are known to every 9th standard student. – Anshul Yadav · 2 years, 8 months ago

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