Initially, a particle is moving at 5.40 m/s at an angle of 37.2° above the horizontal. Two seconds later, its velocity is 6.10 m/s at an angle of 55.0° below the horizontal. What was the particle's average acceleration during these 2.00 seconds in the x-direction and the y-direction

Question

Initially, a particle is moving at 5.40 m/s at an angle of 37.2° above the horizontal. Two seconds later, its velocity is 6.10 m/s at an angle of 55.0° below the horizontal. What was the particle's average acceleration during these 2.00 seconds in the x-direction and the y-direction

Answer 1

To find the average acceleration in the x-direction and the y-direction, we can use the following formulas:

Average acceleration in the x-direction (a_x):
a_x = (v_x2 - v_x1) / t

Average acceleration in the y-direction (a_y):
a_y = (v_y2 - v_y1) / t

Where:
v_x2 and v_x1 are the final and initial velocities in the x-direction,
v_y2 and v_y1 are the final and initial velocities in the y-direction, and
t is the time interval.

Let's calculate the average acceleration in the x-direction first:

v_x2 = 6.10 m/s (final velocity in the x-direction)
v_x1 = 5.40 m/s (initial velocity in the x-direction)
t = 2.00 s (time interval)

a_x = (6.10 m/s - 5.40 m/s) / 2.00 s
= 0.70 m/s / 2.00 s
= 0.35 m/s²

Now, let's calculate the average acceleration in the y-direction:

v_y2 = -6.10 m/s (final velocity in the y-direction, negative because it is below the horizontal)
v_y1 = 5.40 m/s * sin(37.2°) (initial velocity in the y-direction)
= 5.40 m/s * 0.601 (using the trigonometric function sin(37.2°))
= 3.2465 m/s

a_y = (-6.10 m/s - 3.2465 m/s) / 2.00 s
= -9.3465 m/s / 2.00 s
≈ -4.673 m/s²

Therefore, the particle's average acceleration during these 2.00 seconds is approximately 0.35 m/s² in the x-direction and -4.673 m/s² in the y-direction.