Question

a car has a velocity of 10m/s.it acclerates at 0.2m/s for half minute.find the distance travelled during this time and final velocity of car​

Answer 1

$\rm{ \orange{\bigstar Given \longrightarrow }}$

• $\sf{Initial \ velocity = 10m/s}$

• $\sf{Time = 0.5 \ min.}$

• $\sf{Acceleration= 0.2m/s}$

$\rm{ \pink{\bigstar To \ Find \longrightarrow }}$

• $\sf {Distance \ Traveled}$ ?
• $\sf{Final \ velocity}$?

$\rm{ \green{\bigstar Formulas \ used \longrightarrow }}$

• $\boxed{\bf{ s=ut+\dfrac{1}{2}at^2 }}$

• $\boxed{\bf{v^2-u^2=2as}}$

where,

$\sf{s= distance}$

$\sf{u= initial \ velocity}$

$\sf{t= time}$

$\sf{a= acceleration}$

$\sf{v= final \ velocity}$

$\rm{ \blue{\bigstar SoLution \longrightarrow }}$

Firstly we will convert time 0.5 min to sec

we know,

$\sf{1 \ min. = 60 \ sec.}$

$\sf{\implies 0.5 \times 1 min = 0.5 \times 60 sec}$

$\sf{\implies 0.5 min = 30 sec}$

So time [t] = 30 sec.

Now we using newton second equation of motion for finding distance [s],

$\sf{\implies s= ut +\dfrac{1}{2}at^2}$

put given values in the formula we get,

$\sf{\implies s= 10\times 30 + \dfrac{1}{2} \times 0.2 \times 30^2}$

$\sf{ \implies s= 300+ 0.1 \times 900 }$

$\sf{\implies s= 300+90}$

$\boxed{\sf{\bigstar s= 390 m \bigstar }}$

 

Now we will find the value of Final velocity by using newton's third equation,

$\sf{\implies v^2-u^2= 2as}$

put the given values in the formula we get,

$\sf{\implies v^2 - 10^2 = 2\times 0.2 \times 390}$

$\sf {\implies v^2 - 100 = 4\times 39}$

$\sf{\implies v^2 = 156+100}$

$\sf{\implies v^2 = 256}$

$\sf{\implies v= \sqrt{256} }$

$\boxed{\sf{ \bigstar v= 16m/s \bigstar }}$

Therefore,

$\green{\bf{\underline{Distance \ traveled \ by \ car \ is \ 390m.}}}$

$\green{\bf{\underline{And\ Final\ velocity\ of\ car\ is \ 16m/s}}}$.

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Answer 2

Given⟶

\sf{Initial \ velocity = 10m/s}Initial velocity=10m/s

\sf{Time = 0.5 \ min.}Time=0.5 min.

\sf{Acceleration= 0.2m/s}Acceleration=0.2m/s

\rm{ \pink{\bigstar To \ Find \longrightarrow }}★To Find⟶

\sf {Distance \ Traveled}Distance Traveled ?

\sf{Final \ velocity}Final velocity ?

\rm{ \green{\bigstar Formulas \ used \longrightarrow }}★Formulas used⟶

\boxed{\bf{ s=ut+\dfrac{1}{2}at^2 }}s=ut+21at2

\boxed{\bf{v^2-u^2=2as}}v2−u2=2as

where,

\sf{s= distance}s=distance

\sf{u= initial \ velocity}u=initial velocity

\sf{t= time}t=time

\sf{a= acceleration}a=acceleration

\sf{v= final \ velocity}v=final velocity

\rm{ \blue{\bigstar SoLution \longrightarrow }}★SoLution⟶

Firstly we will convert time 0.5 min to sec

we know,

\sf{1 \ min. = 60 \ sec.}1 min.=60 sec.

\sf{\implies 0.5 \times 1 min = 0.5 \times 60 sec}⟹0.5×1min=0.5×60sec

\sf{\implies 0.5 min = 30 sec}⟹0.5min=30sec

So time [t] = 30 sec.

Now we using newton second equation of motion for finding distance [s],

\sf{\implies s= ut +\dfrac{1}{2}at^2}⟹s=ut+21at2

put given values in the formula we get,

\sf{\implies s= 10\times 30 + \dfrac{1}{2} \times 0.2 \times 30^2}⟹s=10×30+21×0.2×302

\sf{ \implies s= 300+ 0.1 \times 900 }⟹s=300+0.1×900

\sf{\implies s= 300+90}⟹s=300+90

\boxed{\sf{\bigstar s= 390 m \bigstar }}★s=390m★

 

Now we will find the value of Final velocity by using newton's third equation,

\sf{\implies v^2-u^2= 2as}⟹v2−u2=2as

put the given values in the formula we get,

\sf{\implies v^2 - 10^2 = 2\times 0.2 \times 390}⟹v2−102=2×0.2×390

\sf {\implies v^2 - 100 = 4\times 39}⟹v2−100=4×39

\sf{\implies v^2 = 156+100}⟹v2=156+100

\sf{\implies v^2 = 256}⟹v2=256

\sf{\implies v= \sqrt{256} }⟹v=256

\boxed{\sf{ \bigstar v= 16m/s \bigstar }}★v=16m/s★

Therefore,

\green{\bf{\underline{Distance \ traveled \ by \ car \ is \ 390m.}}}Distance traveled by car is 390m.

\green{\bf{\underline{And\ Final\ velocity\ of\ car\ is \ 16m/s}}}And Final velocity of car is 16m/s .

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