Final answer:
The usual speed (V) of the plane was found using the average speed formula and equating the adjusted time with the usual time less a 30-minute delay, giving a speed of 750 km/hr, which is option B.
Step-by-step explanation:
The question asks us to find the usual speed of a plane that had to increase its speed by 250 km/hr to cover a distance of 1500 km and arrive on time, despite a 30-minute delay. To calculate this, we will use the formula for average speed, which is Vavg = distance / time.
Let's denote the usual speed of the plane as V, and the increased speed as V + 250 km/hr. The time taken to travel at the usual speed for 1500 km would be 1500 km / V. Because the plane left 30 minutes (or 0.5 hours) late, it must make up that time, which means the actual time for the journey at the increased speed is (1500 km / V) - 0.5 hours. So, we have the equation 1500 km / (V + 250 km/hr) = (1500 km / V) - 0.5 hours.
Now, we solve for V which will require rearranging the equation and solving the resulting quadratic equation. Doing so, we find that the usual speed V is 750 km/hr, which corresponds to option B.
Steps to Solve:
- Set up the equation: 1500 km / (V + 250 km/hr) = (1500 km / V) - 0.5 hours.
- Simplify and solve for V.
- Confirm that the solution matches one of the provided options.