Explanation:
Here,
- Initial velocity (u) = 0 (as it starts from rest)
- Final velocity (v) = 60 km/h
- Time taken (t) = 15 min
Converting the quantities into its standard form :
→ v = 60 km/h
→ v = ( 60 × [tex] \sf \dfrac{5}{18} [/tex] ) m/s
→ v = ( 10 × [tex] \sf \dfrac{5}{3} [/tex] ) m/s
→ v = [tex] \sf \dfrac{50}{3} [/tex] m/s
→ v = 16.66 m/s
Also,
→ t = 15 minutes
→ t = (15 × 60) seconds
→ t = 900 seconds
Calculating acceleration :
★ v = u + at
- v is final velocity
- a is acceleration
- u is initial velocity
- t is time
→ 16.66 = 0 + 900a
→ 16.66 - 0 = 900a
→ 16.66 = 900a
→ [tex] \sf \dfrac{16.66}{900} [/tex] = a
→ 0.0185 m/s² = a
∴ Acceleration is 0.0185 m/s².
_________________________
Calculating distance covered :
★ v² - u² = 2as
- v is final velocity
- a is acceleration
- u is initial velocity
- s is distance
→ (16.66)² - (0)² = 2 × 0.0185 × s
→ 277.5556 - 0 = 0.037s
→ 277.5556 = 0.037s
→ [tex] \sf \dfrac{277.5556}{0.037} [/tex] = s
→ 7501.50 m = s
∴ Distance covered is 7501.50 m .
