Ajet fighter accelerates at 17.7 m/s^2 increasing its velocity from 119 m/s to 233 m/s. how far does it travel in that time

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

A jet fighter accelerates at 17.7 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

Applying the Equation of Motion

substituting the values in above equation, we find

              s= =  

                     =  = 1133.559 m = 1.13 km

Therefore, the distance travelled is 1.13 Km

a 2,000 calorie based diet.