Question

Derive Velocity - time equation by calculus method.

Answer 1

$\textbf{ \large{ \underline{ \: Velocity-time equation : - \: }}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (\textsf{v = u + at})$

Proof - Suppose an object moving with uniform acceleration 'a' if initial velocity of object is 'u' at time 't' = 0 after time 't' final velocity of object is 'v'.

The small change in velocity is dv in time dt. Therefore acceleration will be ,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \text{a} = \frac{ \text{dv}}{ \text{dt}} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \text{a dt = dv} - - - (1)$

Integrating eq (1) over limit of time t = 0 to t from velocity u to v .

$\displaystyle \int\limits_{0}^{ \text{t}}\text{a \: dt }\: = \displaystyle \int\limits_{ \text{u}}^{ \text{v}}\text{ dv} \\ \\ \text{a}\displaystyle \int\limits_{0}^{ \text{t}}\text{ dt } \: = \displaystyle \int\limits_{ \text{u}}^{ \text{v}}\text{ dv} \\ \\ \text{a} \bigg[ \text{t} \bigg] _{0} ^{ \text{t}} \: = \bigg[ \text{v} \bigg] _{u} ^{ \text{v}} \\ \\ \text{a}\bigg[ \text{t - 0} \bigg] \: = \bigg[ \text{v - u} \bigg] \\ \\ \text{at \: = v - u} \\ \\ \rightarrow \: \text{v\: = u + at}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (\textsf{v = u + at})$

Answer 2

Answer:

Derivation of  velocity - time .

⇒ v = u + a t

Also known as first equation of motion .

Step-by-step explanation :

We know change in velocity in unit time is known as acceleration .

i.e.

$\displaystyle{a=\dfrac{dv}{dt} }\\\\\\\displaystyle{a \ dt=dv}$

Integrating both side

$\displaystyle{\int\limits\,a \ dt=dv}\int\limits\, dx$

Applying limit for time and velocity

For velocity ;

Let say lower limit = u

Final limit = v

For time

Let say lower limit = 0

Final limit = t

$\displaystyle{\int\limits^t_0 {a} \,dt=\int\limits^v_u {} \ dv}\\\\\\\displaystyle{a[t]^t_0=[v]^v_u}\\\\\\\displaystyle{at=v-u}\\\\\\\displaystyle{v=u+at}$

Hence , derive velocity - time equation by calculus method.