How to derive kinematics equations using calculus? [closed]

Question

I read derivation of kinematics equations using calculus:

$$a=\frac{\text dv}{\text dt}$$ $$\implies \text dv=a\text dt$$ $$\implies \int_{v_0}^v\text dv=\int_0^t a\text dt$$ $$\implies v-v_0=at$$ $$\implies v=v_0+at\tag1$$

I know finding antiderivatives and basic concepts of integration:

I cannot understand:

1. How can we take $dt$ in the first equation to the other side when $dv/dt$ is not a fraction?

2. In third equation how have we placed the upper and lower limits in LHS of velocity and in RHS of time?

3. In third equation we have only $dv$ in LHS. Then what will be the function we are integrating?

Lastly can you please suggest some websites from where I can learn how to integrate both sides of an equation like done above.

Please provide me the answers.

Answer 1

1. Since the RHS in your first expression is a total derivative, it is OK to move the differentials around in this way.
2. $v_0\equiv v(t=0)$ and $v\equiv v(t)$, i.e., the velocity limits of integration are just the velocities evaluated at the time limits of integration. Note that the velocity integral is abusing notation a bit, since the integration variable also appears as a limit of integration. A slightly better way to write this integral would be $$\int_{v_0}^vdv',$$ introducing the "dummy variable" $v'$ to act as the integration variable.
3. The integrand on the LHS is 1 :)

Answer 2

1) By error approximation we can show that $\frac{dy}{dx}$ is actually the ratio of dy and dx.

2)In the third equation acceleration a is constant, and the variables are velocity and time. SO we integrate with respect to their corresponding limits.

3) $dv = 1.dv$ . So the integrand is unity i.e, 1.

Answer 3

1. $\dfrac{dv}{dt}$ is not a fraction, but it is the limit of $\dfrac{\Delta v}{\Delta t}$ which is a fraction. The trick here is to do the algebraic manipulations before taking the limit.

2. The LHS integral is over the variable $v$, while the RHS is over the variable $t$, so the integration interval endpoints must be expressed in terms of values for the corresponding variable.

3. The function is the constant function $1$ in this case: $dv = 1\cdot dv$.

The Wikipedia article on Infinitesimal has good information on the mathematical issues.