Question

A particle moving in the x-y plane has a velocity at time t=6 s given by 4i+5j m/s, and at t = 6.1 s its velocity has become 4.3i+5.4j m/s. Calculate the magnitude aav of its average acceleration during the 0.1-s interval and the angle theta it makes with the x-axis.

Answer 1

The magnitude of the average acceleration during the 0.1-s interval is 5 m/s² and it makes an angle of approximately 53.13° with the x-axis.

Step-by-step explanation:

To find the magnitude of the average acceleration, we can use the formula:
aav = (vf - vi) / t

Given that the velocity at t = 6s is 4i + 5j m/s, and at t = 6.1s it is 4.3i + 5.4j m/s, we can calculate the change in velocity: Δv = (4.3i + 5.4j) - (4i + 5j) = 0.3i + 0.4j

Since the time interval is 0.1s, we can substitute the values into the formula:
aav = (0.3i + 0.4j) / 0.1 = 3i + 4j m/s²

To find the angle θ the average acceleration makes with the x-axis, we can use the formula:
θ = atan2(aav.y, aav.x)

Substituting the values we obtained:
θ = atan2(4, 3) ≈ 53.13°