Question

A student enters a parking lot and walks toward the north at a velocity of 1.20 m/s for 50.52 seconds. He then turns to the west and runs at a pace of 3.56 m/s for 25.44 seconds. What is the total displacement of the student?

A. 78.0 m
B. 105.6 m
C. 131.2 m
D. 143.2 m

Answer 1

The total displacement of the student is 109.063 m.

Step-by-step explanation:

To find the total displacement of the student, we need to calculate the individual displacements in the north and west directions. The north displacement can be calculated as the product of the velocity and time, which is 1.20 m/s * 50.52 s = 60.624 m. The west displacement can also be calculated similarly, which gives 3.56 m/s * 25.44 s = 90.5024 m.

To find the total displacement, we can use the Pythagorean theorem to calculate the magnitude. So, the magnitude of the total displacement is sqrt((60.624 m)² + (90.5024 m)²) = 109.063 m.