Question

Port A is defined to be the origin of a set of coordinate axes and port B is located at the point (70,30), where the distances are measured in kilometres. A ship S1​ sails from port A at 10:00 in a straight line such that its position t hours after 10:00 is given by r=t(1020​). A speedboat S2​ is capable of three times the speed of S1​ and is to meet S1​ by travelling the shortest distance possible. What is the latest time that S2​ can leave port B?

Answer 1

To determine the latest time that S2 can leave port B, we need to compute the distance from Port A to Port B, determine the speed of S2, and finally use the formula Time = Distance / Speed. Approximately, S2 must leave port B not later than 10:01:30.

Step-by-step explanation:

The problem at hand revolves around speed, time and distance, typically dealt with in Mathematics, more precisely in Geometry and Algebra. Given the information, we will first determine the speed of S1. S1 was described as moving at a rate of (t(1020)), meaning it was moving 1020 km per hour.

Now, S2's speed is described as three times that of S1, so S2 moves at a speed of 1020 * 3 = 3060 km/hour. The distance from Port A to Port B is the Euclidean distance which can be calculated from the coordinates, which is approximately 76.81 km.

To find the latest possible time for S2 to leave, we will solve for time on the formula Time = Distance / Speed. So, Time = 76.81 / 3060 = 0.025 hours, which is approximately 1.5 minutes. That means S2 must leave port B no later than 10:01:30 to meet ship S1.

Learn more about Speed, Distance and Time