A cyclist cycles 50.0 km [N] and then 30.0 km [E]. The total time taken for the trip is 1.00 h. What is its average velocity?
Question 7 options:
A)
3.40 × 10 3 km/h [31° E of N]
B)
58.3 km/h [31° E of N]
C)
3.40 × 10 3 km/h [59° E of N]
D)
58.3 km/h [59° E of N]
E)
58.3 km/h [NE ]
We need to find the total displacement and divide it by the total time to get the average velocity. Using the Pythagorean theorem, the displacement is:
d = sqrt((50.0 km)^2 + (30.0 km)^2) = 58.3 km
The direction can be found using trigonometry:
theta = arctan(30.0 km / 50.0 km) = 31° E of N
Therefore, the average velocity is:
v = d / t = 58.3 km / 1.00 h = 58.3 km/h [31° E of N]
The answer is (B).