Question

Geoffrey drives from his home town due east at \(90.0 \mathrm{km} / \mathrm{h}\) for 80.0 min. After visiting a friend for 15.0 min, he drives in a direction \(30.0^{\circ}\) south of west at \(76.0 \mathrm{km} / \mathrm{h}\) for 45.0 min to visit another friend. (a) How far is it to his home from the second town? (b) If it takes him 45.0 min to drive directly home, what is his average velocity on the third leg of the trip? (c) What is his average velocity during the first two legs of his trip? (d) What is his average velocity over the entire trip? (e) What is his average speed during the entire trip if he spent 55.0 min visiting the second friend?

Short answer

Short answer: After a detailed analysis of the trip, the answers for each requested scenario are as follows: (a) The distance to the second town is 104.07 km. (b) The average velocity on the third leg of the trip is 138.76 km/h. (c) The average velocity for the first two legs of the trip is 43.45 km/h (east-west) and 23.47 km/h (north-south). (d) The average velocity for the entire trip is 29.67 km/h (east-west) and 16.00 km/h (north-south) (e) The average speed during the entire trip, including the time spent visiting the second friend, is 70.27 km/h.

Step-by-step solution

Convert time in minutes to hours

Divide the given time in minutes by 60 to convert them into hours. 80.0 min: 80.0 / 60 = 1.333 h 45.0 min: 45.0 / 60 = 0.75 h 15.0 min: 15.0 / 60 = 0.250 h 45.0 min: 45.0 / 60 = 0.75 h 55.0 min: 55.0 / 60 = 0.917 h

Compute displacement for each leg of the trip

Multiply the time (in hours) by the speed of each leg to find the corresponding displacement: Displacement for leg 1 (due East): 90.0 km/h * 1.333 h = 120 km (east) Displacement for leg 2 (30° south of west): 76.0 km/h * 0.75 h = 57 km (30° south of west)

Find components of the displacements for leg 2

We have to split the displacements into East-West and North-South components. East component of displacement for leg 2: 57 km * sin(30) = 28.5 km (west) South component of displacement for leg 2: 57 km * cos(30) = 49.34 km (south)

Compute the total displacement for the first two legs

Add the components of the displacements found in Steps 2 and 3 to find the total displacement: Total East-West displacement: 120 km (east) - 28.5 km (west) = 91.5 km (east) Total North-South displacement: 49.34 km (south)

Compute the distance to the second town (a)

Use the Pythagorean theorem to find the distance between the starting point and the second town: Distance = sqrt((91.5 km)^2 + (49.34 km)^2) = 104.07 km

Compute the average velocity on the third leg (b)

Using the previously derived distance and the given time of 45.0 minutes (0.75 hours): Average velocity on third leg = (104.07 km) / (0.75 h) = 138.76 km/h

Compute the average velocity for the first two legs (c)

Add the total displacements of the first two legs and divide by the total time: Average velocity (east-west) = (91.5 km) / (1.333 h + 0.75 h) = 43.45 km/h Average velocity (north-south) = (49.34 km) / (1.333 h + 0.75 h) = 23.47 km/h

Compute the average velocity over the entire trip (d)

Add the total displacements of all legs and divide by the total time. Note that the third leg brings Geoffrey back to the starting point, so the total east-west and north-south displacements remain as 91.5 km and 49.34 km, respectively: Total time = 1.333 h + 0.250 h + 0.75 h + 0.75 h = 3.083 h Average velocity (east-west) = (91.5 km) / (3.083 h) = 29.67 km/h Average velocity (north-south) = (49.34 km) / (3.083 h) = 16.00 km/h

Compute the average speed during the entire trip (e)

Calculate the total distance travelled and divide by the total time, including the time spent visiting the second friend: Total distance = 120 km (leg 1) + 57 km (leg 2) + 104.07 km (leg 3) = 281.07 km Total time = 1.333 h + 0.250 h + 0.75 h + 0.75 h + 0.917 h = 4.000 h Average speed = (281.07 km) / (4.00 h) = 70.27 km/h