Acceleration (a) = 5.1 m/s²
Initial speed (u) = 16 km/h ⇒ [tex] \frac{16*1000}{3600} [/tex] m/s ≈ 4.5m/s
Final speed (v) =118 km/h ⇒ [tex] \frac{118*1000}{3600} [/tex] m/s ≈ 32.8m/s
Distance(S) travel in that particular instant is carried out by 'Third equation of motion' i.e., v² = u² + 2aS
So, When all quantities are in S.I. unit then,
putting the values in the equation of motion,
As we have to carry out the distance covered,
2·a·S = v² - u²
S = [tex] \frac{ v^{2} - u^{2} }{2a} [/tex]
Putting values in derived equation,
⇒ S = [tex] \frac{ (32.8)^{2} - (4.5)^{2} }{2*(5.1)} [/tex]
⇒ S = [tex] \frac{ 1075.84 - 20.25}{10.2} [/tex]
⇒ S = [tex] \frac{1055.59}{10.2} [/tex]
⇒ S ≈ 103.489 m
The total distance covered in that given condition is approx. 103.289 m.
