Question

If 5 m/s and 10 m/s are the initial and final velocities of a body having a uniform acceleration in same time interval. What will be its average velocity? What will be the average velocity if the motion is not uniform?

Answer 1

Answer:

Given:

Initial Velocity = 5 m/s

Final Velocity = 10 m/s

Acceleration is constant

To find:

Average velocity in that journey

Concept:

Average speed is always defined as the ratio of total distance to the total time taken .

Let averaging speed be avg. v

$avg. \: v = \dfrac{distance}{time}$

$= > avg. \: v = \dfrac{s}{t}$

$= > avg. \: v = \dfrac{ut + \frac{1}{2} a {t}^{2} }{t}$

$= > avg. \: v = u + \frac{1}{2} at$

Putting value of "a" :

$= > avg. \: v = u + \dfrac{1}{2} \bigg(\dfrac{v - u}{t} \bigg) t$

$= > avg. \: v = \dfrac{u + v}{2}$

Calculation:

Putting all the available value , we get :

$avg. \: v = \dfrac{u + v}{2}$

$= > avg. \: v = \dfrac{(5 + 10)}{2}$

$= > avg. \: v = \dfrac{(15)}{2}$

$= > avg. \: v = 7.5 \: m {s}^{ - 1}$

So final answer :

$\boxed{ \red{ \huge{ \bold{avg. \: v = 7.5 \: m {s}^{ - 1} }}}}$

Answer 2

GiveN :

• Initial velocity (u) = 5 m/s
• Final velocity (v) = 10 m/s

To FinD :

• Average Velocity

SolutioN :

Use formula for Average velocity :

$\dashrightarrow \boxed{\tt{V_{avg} \: = \: \dfrac{v \: + \: u}{2}}} \\ \\ \dashrightarrow \tt{V_{avg} \: = \: \dfrac{5 \: + \: 10}{2}} \\ \\ \dashrightarrow \tt{V_{avg} \: = \: \dfrac{15}{2}} \\ \\ \dashrightarrow \tt{V_{avg} \: = \: 7.5} \\ \\ \underline{\boxed{\bf{Average \: velocity \: is \: 7.5 \: ms^{-1}}}}$

$\rule{150}{0.5}$

AdditionaL InformatioN :

• Velocity is defined as rate of change of displacement.

• Velocity is a vector quantity.

• Velocity is denoted by v or u

• It's SI unit is m/s