Question

Jason is running with an initial velocity of 7.2 m/s [N]. He swerves to avoid a tackle, and after 2.0 s, is moving at 7.2 m/s [W]. His average acceleration over the time interval is: ________

Answer 1

Jason's average acceleration, after changing his direction from north to west, is 3.6√2 m/s² in a northwest direction.

Step-by-step explanation:

Jason's average acceleration can be determined by the change in his velocity vector over the time interval. Since velocity is a vector quantity, we need to consider the directional change. Initially, Jason is running north at 7.2 m/s and then west at 7.2 m/s.

The velocity change is thus the vector from the initial to the final velocity. Using Pythagoras theorem, the magnitude of this change in velocity (Δv) is √(7.2² + 7.2²) m/s, which simplifies to 7.2√2 m/s. The average acceleration, a, is the change in velocity (Δv) divided by the time interval (Δt), which is 2.0 s.

Δv = 7.2√2 m/s
a = Δv / Δt = (7.2√2 m/s) / 2.0 s = 7.2√2 / 2 m/s² = 3.6√2 m/s²

The direction of this acceleration will be northwest because it's the direction from 'north' towards 'west' considering we are taking north as positive y-axis and west as negative x-axis.