Final answer:
Using the kinematic equation, the car's final velocity after accelerating over a distance of 0.25 km approximates to its original speed of 20 km/h. The increase in velocity due to the acceleration is negligible over this short distance.
Step-by-step explanation:
To calculate the car's final velocity after it has covered a distance of 0.25 km under constant acceleration, we use the kinematic equation v = √(v₀² + 2ad), where v is the final velocity, v₀ is the initial velocity, a is the acceleration, and d is the distance covered.
Given:
• Initial velocity, v₀ = 20 km/h
• Acceleration, a = 5 km/h/s (which needs to be converted to km/h² for consistency in units)
• Distance, d = 0.25 km
First, convert the acceleration to km/h².
• 1 km/h/s = 1 km/h * (1/3600 s)−¹ = 1/3600 km/h²
• a = 5 km/h/s * (1/3600 s)−¹ = 5/3600 km/h²
Now, apply the formula:
• v = √(20² + 2 * (5/3600) * 0.25)
• v = √(400 + 2 * (5/3600) * 0.25)
• v = √(400 + 0.000694444)
• v = √(400.000694444)
• v ≈ 20.000017 km/h
After rounding to a reasonable level of precision, we can conclude that the car's velocity approximates to its original speed of 20 km/h since the increase due to the given acceleration over the distance of 0.25 km is negligible.