Question

What is the formula for time from a changing velocity?

Answer 1

Hint: To solve the question we need to know the concept of velocity. Velocity is defined as the rate of change of displacement. Acceleration is the ratio of rate of change in velocity which means it is the ratio of change in velocity to change in time.

Complete step by step answer:
The question asks us to find the formula for time from a changing velocity. To do this the first step is to write the acceleration in terms of velocity. Acceleration is the ratio of change in velocity to change in time. So acceleration which is constant is written as:
$a=\dfrac{\vartriangle v}{\vartriangle t}$
$\Rightarrow a=\dfrac{{{v}_{2}}-{{v}_{1}}}{{{t}_{2}}-{{t}_{1}}}$
Considering the time ${{t}_{1}}=0$ and ${{t}_{2}}=t$ we get:
$\Rightarrow \text{a =}\dfrac{{{\text{v}}_{\text{2}}}\text{-}{{\text{v}}_{\text{1}}}}{\text{t}}$
$\Rightarrow \text{t =}\dfrac{{{\text{v}}_{\text{2}}}\text{- }{{\text{v}}_{\text{1}}}}{\text{a}}$
To re-write the formula of time in a much better way, we can take help of the other formula which is in terms of displacement, acceleration and time. The formula is:
$\text{s = }{{\text{v}}_{1}}\text{t +}\dfrac{\text{1}}{\text{2}}\text{a}{{\text{t}}^{\text{2}}}$
Here ${{u}_{0}}$ is the initial speed, $t$ is the time and $a$is the acceleration. So time in terms of velocity and acceleration could be written as:
$\text{s = }{{\text{v}}_{1}}\text{t +}\dfrac{\text{1}}{\text{2}}\text{a}{{\text{t}}^{\text{2}}}$
Substituting the value of acceleration with $a=\dfrac{{{v}_{2}}-{{v}_{1}}}{t}$the above formula changes to:
$\Rightarrow \text{s = }{{\text{v}}_{1}}\text{t +}\dfrac{\text{1}}{\text{2}}\left( \dfrac{{{\text{v}}_{\text{2}}}\text{-}{{\text{v}}_{\text{1}}}}{\text{t}} \right){{\text{t}}^{\text{2}}}$
The time $''t''$in denominator cancels as in the numerator there is presence of ${{t}^{2}}$.
$\Rightarrow \text{s = }{{\text{v}}_{1}}\text{t +}\dfrac{\text{1}}{\text{2}}\left( {{\text{v}}_{\text{2}}}\text{- }{{\text{v}}_{\text{1}}} \right)\text{t}$
Taking the time $t$ common we get
$\Rightarrow \text{s = t}\left( {{\text{v}}_{\text{1}}}\text{ +}\dfrac{\text{1}}{\text{2}}\left( {{\text{v}}_{\text{2}}}\text{- }{{\text{v}}_{\text{1}}} \right) \right)$
Taking the whole term of velocity from L.H.S. to R.H.S. the term divides $\text{''s''}$ which is displacement, we get:
$\Rightarrow \text{t = }\dfrac{\text{s}}{{{\text{v}}_{\text{1}}}\text{ +}\dfrac{\text{1}}{\text{2}}\left( {{\text{v}}_{\text{2}}}\text{- }{{\text{v}}_{\text{1}}} \right)}$
$\therefore $ The formula for time from a changing velocity is $\text{t = }\dfrac{\text{s}}{{{\text{v}}_{\text{1}}}\text{ +}\dfrac{\text{1}}{\text{2}}\left( {{\text{v}}_{\text{2}}}\text{- }{{\text{v}}_{\text{1}}} \right)}$.

Note: The above formula could be used to find the time taken by the object to travel a certain distance with the given velocity. The above question is valid for the constant acceleration. If the acceleration is the function of time like $a=f\left( v \right)$ then then the formula for time is $t=\int\limits_{{{v}_{1}}}^{{{v}_{2}}}{\dfrac{dv}{a\left( v \right)}}$ .