Question

A particle starts its motion with initial velocity u and moves with uniform acceleration if its velocity is v after time t then find out distance travelled if in this duration?​

Answer 1

Answer:

$\frac{t(v + u)}{2}$

Explanation:

Given

Initial velocity = 'u'

Final Velocity = 'v'

Time = 't'

Distance = 'S'

We Know That

=> v = u + at

=> v - u = at

$= > \boxed{ a = \frac{v - u}{t} }$

By second Equation of Motion

$= > \boxed{S = ut + \frac{1}{2} a {t}^{2} }$

On Putting Given Values

$= > S = ut + \frac{1}{2} ( \frac{v - u}{t} ) \times {t}^{2} \\ = > S = ut + \frac{1}{2} \frac{(v - u)}{ \cancel{t}} \times \cancel{t} \times t$

$= > S = ut + \frac{vt - ut}{2}$

On Taking L.C.M.

$= > S = \frac{2ut + vt - ut}{2}$

$= > S = \frac{vt + ut }{2}$

On taking 't' as Common

$= > \boxed{S = \frac{t(v + u)}{2} }$

Answer 2

• Initial velocity (during starting) = u

• Final velocity (at last) = v

• Time = t

_______________ [ GIVEN ]

• We have to find the distance travelled by it.

____________________________

Now..

$\boxed{*\:v\: =\: u\: +\: at}$

(From first law of Motion)

→ v = u + at

→ v - u = at

→ at = v - u

→ a = $\dfrac{v\:-\:u}{t}$ ______ (eq 1)

___________________________

We have to find the distance (s).

So..

We know that that

$\boxed{s \:= \:ut \:+\: \frac{1}{2}a{t}^{2}}$

(Second law of Motion)

Put value of a in above formula

→ s = ut + $\dfrac{1}{2}$ × $\dfrac{v\:-\:u}{t}$ × t²

→ s = ut + $\dfrac{1}{2}$ (v - u)t

→ s = ut + $\dfrac{(v\:-\:u)t}{2}$

→ s = $\dfrac{2ut\:+\:(v\:-\:u)t}{2}$

→ s = $\dfrac{2ut\:+\:vt\:-\:ut}{2}$

→ s = $\dfrac{ut\:+\:vt}{2}$

→ s = $\dfrac{t(u\:+\:v)}{2}$

____________________________

s = $\dfrac{t(u\:+\:v)}{2}$

_________ [ ANSWER ]

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