Answer:
1130 m
Explanation:
Given:
a = 17.7 m/s²
v₀ = 119 m/s
v = 233 m/s
Find: Δx
v² = v₀² + 2aΔx
(233 m/s)² = (119 m/s)² + 2(17.7 m/s²) Δx
Δx ≈ 1130 m
Answer:
A jet fighter accelerates at 17.7[tex]m/s^2[/tex] increasing its velocity from 119 m/s to 233 m/s. The distance that it can travel in that time is 1133.559 m or 1.13 km
Explanation :
From the Given statements, we know that
Initial Velocity (u) = 119 m/s
Final Velocity (v) = 233 m/s
Acceleration (a) = 17.7 [tex]m/s^2[/tex]
Applying the Equation of Motion
[tex]\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{as}[/tex]
substituting the values in above equation, we find
s= [tex]\frac{v^{2}-u^{2}}{2 a}[/tex] = [tex]\frac{(233)^{2}-(119)^{2}}{2 \times 17.7}[/tex]
=[tex]\frac{54289-14161}{35.4}[/tex] = 1133.559 m = 1.13 km
Therefore, the distance travelled is 1.13 Km
