Jason drives due west with a speed of 55.0km/hr for 30.0 min, then continues in the same direction with a speed of 95.0km/hr for 2.00 hr, then drives farther west at 40.0km/hr for 10.0 min. What is the magnitude of Jason's average velocity
Some sliding rocks approach the base of a hill with a speed of 16.0 m/s. The hill rises at 38.0^\circ above the horizontal and has coefficients of kinetic and static friction of 0.400 and 0.620, respectively, with these rocks. Start each part of your solution to this problem with a free-body diagram.
To determine the magnitude of Jason's average velocity, we need to calculate the total displacement and divide it by the total time taken.
First, let's calculate the displacement for each segment of the journey:
1. In the first segment, Jason drives west for 30.0 minutes at a speed of 55.0 km/hr. The time needs to be converted to hours: 30.0 min = 30.0/60 = 0.5 hr. The displacement for this segment is:
Distance = Speed * Time = 55.0 km/hr * 0.5 hr = 27.5 km west.
2. In the second segment, Jason continues driving west at a higher speed of 95.0 km/hr for 2.00 hours. The displacement for this segment is:
Distance = Speed * Time = 95.0 km/hr * 2.00 hr = 190.0 km west.
3. In the third segment, Jason drives west for another 10.0 minutes at a speed of 40.0 km/hr. Converting the time to hours: 10.0 min = 10.0/60 = 0.167 hr. The displacement for this segment is:
Distance = Speed * Time = 40.0 km/hr * 0.167 hr = 6.67 km west.
Now, let's calculate the total displacement by adding up the displacements from all three segments:
Total displacement = 27.5 km + 190.0 km + 6.67 km = 224.17 km west.
To find the average velocity, we divide the total displacement by the total time taken:
Total time = 0.5 hr + 2.00 hr + 0.167 hr = 2.667 hr.
Average velocity = Total displacement / Total time = 224.17 km / 2.667 hr = 84.02 km/hr.
Therefore, the magnitude of Jason's average velocity is 84.02 km/hr.