<em>Acceleration (a) = 5.1 m/s</em>²
<em>Initial speed (u) = 16 km/h </em>⇒
<em>m/s </em>≈ <em>4.5m/s</em><em>
</em><em>Final speed (v) =118 km/h </em>⇒
<em>m/s</em> ≈ <em>32.8m/s</em><em>
</em>
Distance(S) travel in that particular instant is carried out by 'Third equation of motion' i.e., v² = u² + 2aS
<em>So, When all quantities are in S.I. unit then,
</em>putting the values in the equation of motion,
<em /><em>As we have to carry out the distance covered,
</em><em>2</em>·<em>a</em>·<em>S = v</em>² <em>- u</em>²
<em>S = </em>
Putting values in derived equation,
⇒ <em>S = </em>
⇒ <em>S = </em>
⇒ <em>S = </em>
⇒ <em>S </em>≈ <em>103.489</em> <em>m
</em><em>
The total distance covered in that given condition is approx. 103.289 m.</em>
<em>Initial speed (u) = 16 km/h </em>⇒
</em><em>Final speed (v) =118 km/h </em>⇒
</em>
Distance(S) travel in that particular instant is carried out by 'Third equation of motion' i.e., v² = u² + 2aS
<em>So, When all quantities are in S.I. unit then,
</em>putting the values in the equation of motion,
<em /><em>As we have to carry out the distance covered,
</em><em>2</em>·<em>a</em>·<em>S = v</em>² <em>- u</em>²
<em>S = </em>
Putting values in derived equation,
⇒ <em>S = </em>
⇒ <em>S = </em>
⇒ <em>S = </em>
⇒ <em>S </em>≈ <em>103.489</em> <em>m
</em><em>
The total distance covered in that given condition is approx. 103.289 m.</em>