Question

If a body starts from rest and accelerates to a speed of 5 m/s in 1 second, then the distance travelled by it is​

Answer 1

Answer

Given -

$\bf u = 0 m/s$

$\bf v = 5 m/s$

$\bf t = 1 sec$

where

$\longrightarrow$u is initial velocity.

$\longrightarrow$v is final velocity.

$\longrightarrow$t is time taken.

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To find -

Distance travelled $\longrightarrow$ s

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Solution -

By substituting the value in 1st equation of motion , first we will find acceleration. After finding acceleration we can substitute the values in both 2nd and 3rd equation of motion.

$\longrightarrow$$\bf u = 0 m/s$

$\longrightarrow$$\bf v = 5 m/s$

$\longrightarrow$$\bf t = 1 sec$

Substituting the value in 1st equation of motion -

$\implies$$\bf v = u + at$

$\implies$$\bf 5 = 0 + a$

$\implies$$\bf a = 5m/s^2$

Acceleration of particle is $\bf 5 m/s^2$

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$\longrightarrow$$\bf u = 0 m/s$

$\longrightarrow$$\bf t = 1 sec$

$\longrightarrow$$\bf a = 5 m/s^2$

Substituting the value in 2nd equation of motion -

$\implies$$\bf s = ut + 1/2 at^2$

$\implies$$\bf s = 0 + 1/2 \times 5 \times 1^2$

$\implies$$\bf s = 1/2 \times 5$

$\implies$$\bf s = 2.5 m$

Distance travelled in 1sec is 2.5 m

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Answer 2

$\huge\rm { _!! Question !_! }$

If a body starts from rest and accelerates to a speed of 5 m/s in 1 second, then the distance travelled by it is?

$\huge\rm { _!! Solution !_! }$

${\underline {\underline {\rm {\red { Given: }}}}}$

• $\rm\pink { (u ) Initial \: velocity = 0m/s }$

$\rm { (as \: it \: started \: from \: rest .) }$

• $\rm\pink { (v ) Final \: velocity = 5m/s }$

• $\rm\pink { (t ) time \: taken = 1s }$

${\underline {\underline {\rm {\red { To \: find : }}}}}$

• $\rm\blue { ( s ) total \: distance \: travelled }$

${\underline {\underline {\rm {\red { Calculation : }}}}}$

We can find total distance by 2nd equation of motion-

$\rm\red { \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: s = ut + \dfrac{1}{2}a{t}^{2} }$

Where,

• $\rm { s = distance ( need \: to \: find )}$

• $\rm { u = initial \: velocity ( 0m/s )}$

• $\rm { v = final \: velocity ( 5m/s )}$

• $\rm { a = acceleration}$

• $\rm { t = time ( 1s )}$

But there is not acceleration given ,so we need to find acceleration first,

$\rm{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: a = \dfrac{v-u}{t} }$

$\rm{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: a = \dfrac{5-0}{1} }$

$\rm{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: a = 5m/s }$

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Now we have,

• $\rm\purple{ (u ) Initial \: velocity = 0m/s }$

• $\rm\purple { (v ) Final \: velocity = 5m/s }$

• $\rm\purple { (t ) time \: taken = 1s }$

• $\rm\purple { (a ) acceleration = 5m/s }$

$\rm\blue { \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: s = ut + \dfrac{1}{2}a{t}^{2} }$

$\rm { \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \leadsto s =( 0 \times 1 )+ \dfrac{1}{2} \times 5 \times {1}^{2} }$

$\rm { \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \leadsto s = \dfrac{5}{2} \times 1 }$

$\rm\orange { \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \leadsto s = \cancel { \dfrac{5}{2}} = 2.5m }$

Therefore, distance travelled by it is 2.5m.

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$\huge\rm { _!! Additional \: information !_! }$

Three equations of motion:

• ${\underline {\underline {\rm {\red { v = u + at }}}}}$

• ${\underline {\underline {\rm {\red { s = ut + \dfrac{1}{2}a{t}^{2} }}}}}$

• ${\underline {\underline {\rm {\red { 2as = {v}^{2} - {u}^{2} }}}}}$

Where,

• v is final velocity
• u is intial velocity
• a is acceleration
• t is time
• s is distance/displacement

To solve problems related on motion we should remember that -

• If a body starts from rest or stationary position then $\rm { u = 0 }$

• If a body comes rest ( it stops ) or breaks are applied then $\rm { v= 0 }$

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