Question

A body initially at rest travels a distance of 100 m in 5 see with a constant acceleration calculate i. Acceleration ii. Final velocity at end of 5sec.2​

Answer 1

$\large \dag$ Question :-

A body initially at rest travels a distance of 100 m in 5 sec with a constant acceleration calculate

(i) Acceleration

(ii) Final velocity at end of 5 sec.

$\large \dag$ Answer :-

$\red\dashrightarrow\underline{\underline{\sf  \green{Acceleration   \:  is \:  8\: m/s^2 }} }$

$\red\dashrightarrow\underline{\underline{\sf  \green{Final\:Velocity   \:  is \:  40\: m/s}} }$

$\large \dag$ Step by step Explanation :-

❒ We Have 2nd Equation of Motion as :

Here We Have :

• Initial Velocity = u = 0 m/s

• Distance Covered = s = 100 m

• Time Taken = t = 5 s

• Let Acceleration be = a m/s²

• Let Final Velocity be = v m/s

$\large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{   \blue{ s = ut + \frac{1}{2} {at}^{2} }}}}$

⏩ Applying 2nd Equation of Motion ;

$:\longmapsto \rm 100 = 0 \times t + \frac{1}{2} \times a \times {t}^{2}$

$:\longmapsto \rm 100 = 0 + \frac{1}{2} \times a \times 25$

$:\longmapsto \rm 100 = \frac{25a}{2}$

$:\longmapsto \rm a = 100 \div \frac{25}{2}$

$:\longmapsto \rm a = 100 \times \frac{2}{25}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf a = 8} }}}$

Therefore,

$\large\underline{\pink{\underline{\frak{\pmb{\text Acceleration = 8 \: m/s^2 }}}}}$

Now

❒ We Have 1st Equation of Motion as :

$\large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{   \blue{v = u+ at  }}}}$

⏩ Applying 1st Equation of Motion ;

$:\longmapsto \rm v = 0 + 8 \times 5$

$:\longmapsto \rm v = 0 + 40$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf v = 40} }}}$

Therefore,

$\large\underline{\pink{\underline{\frak{\pmb{Final \: \text Velocit \text y = 40 \: m/s }}}}}$

Answer 2

1. Acceleration

Using formula of 2nd eq. of motion

$\sf\large\red{\underline{\underline {\underline{s = ut + \frac{1}{2}at^2}}}}$

Where,

s is 100m

t is 5sec

u is 0m/s

On putting these values in the formula, we get,

$\sf \implies \: \: {100 = 0 \times t + \frac{1}{2} \times a \times {t}^{2}}$

$\sf \implies \: \: {100 = 0 + \frac{1}{2} \times a \times {5}^{2} }$

$\sf \implies\:\:{100 = 0 + \frac{1}{2} \times a \times 25}$

$\sf \implies \: \: {100 = \frac{25}{2}a}$

$\sf \implies \: \: {a = 100 \times \bigg(\frac{2}{25}\bigg)}$

$\sf \implies \: \: {a = \frac{100 \times 2}{25} = {\frak{\red{\underline{\underline{8m/s^2}}}}}}$

2. Final Velocity

Using formula of 3rd eq. of motion

$\sf\large\red{\underline{\underline {\underline{ {v}^{2} = {u}^{2} + 2as}}}}$

On putting values, we have,

$\sf \implies \: \: { {v}^{2} = {0}^{2} + 2 \times 8 \times 100}$

$\sf \implies \: \: { {v}^{2} = 0 + 2 \times 8 \times 100}$

$\sf \implies \: \: { {v}^{2} = 1600}$

$\sf \large \red{\underline{\underline{\implies \: \: { {v} = 40m/s}}}}$