Carlo drives his car for a picnic by first going 105 km north to pick up his girlfriend and then 120.0 km east to reach the picnic groove. The whole trip takes 135 min. Calculate (a) Carlo’s average speed and (b) average velocity.
135 min = 2.3 hr
speed = (1/2.3) sqrt (105^2+120^2) km/hr
velocity = speed at angle tan^-1(120/105) east of north
To calculate Carlo's average speed, we need to use the formula:
Average Speed = Total Distance / Total Time
(a) Calculating average speed:
1. Calculate the total distance covered by Carlo: The distance going north and east can be treated as the two sides of a right-angled triangle. We can use the Pythagorean theorem to calculate the distance.
Distance = √[(105 km)^2 + (120.0 km)^2]
= √[11025 km^2 + 14400 km^2]
= √(25425 km^2)
≈ 159.508 km
2. Convert the total time to hours: Since the given time is in minutes, we need to convert it to hours. There are 60 minutes in an hour.
Total Time = 135 min / 60
= 2.25 hours
3. Calculate the average speed:
Average Speed = Total Distance / Total Time
= 159.508 km / 2.25 hours
≈ 70.893 km/h
Therefore, Carlo's average speed is approximately 70.893 km/h.
(b) Calculating average velocity:
Carlo's average velocity not only considers the distance traveled but also takes into account the displacement, which is the straight-line distance from the starting point to the end point (regardless of path).
1. Calculate the displacement: The displacement is the straight-line distance between the starting point and the end point of the trip. It can be calculated using the Pythagorean theorem for the right-angled triangle formed by the north and east distances.
Displacement = √[(105 km)^2 + (120.0 km)^2]
= √[11025 km^2 + 14400 km^2]
= √(25425 km^2)
≈ 159.508 km
2. Calculate the average velocity:
Average Velocity = Displacement / Total Time
= 159.508 km / 2.25 hours
≈ 70.893 km/h
Therefore, Carlo's average velocity is approximately 70.893 km/h.
Note: In this case, the average speed and average velocity are the same because the direction of motion (north and east) is along a straight line.