A particle travels to the right at a constant

rate of 6.5 m/s. It suddenly is given a vertical
acceleration of 3.2 m/s2 for 3.9 s.
What is its direction of travel after the
acceleration with respect to the horizontal?
Answer between −180� and +180�.
Answer in units of �.

What is the speed at this time?
Answer in units of m/s.

What is your thinking on this? I will be happy to check your work.

To determine the direction of travel after the acceleration with respect to the horizontal, we need to break down the motion into horizontal and vertical components.

Given that the particle travels to the right at a constant rate of 6.5 m/s, the horizontal component of its motion remains unchanged after the acceleration. Therefore, its horizontal direction of travel remains to the right.

To determine the vertical direction of travel after the acceleration, we consider the vertical acceleration of 3.2 m/s^2 for 3.9 seconds. Using the kinematic equation:

vf = vi + at

where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time, we can find the vertical component of the final velocity.

vf = 0 + (3.2 m/s^2)(3.9 s) = 12.48 m/s

The speed at this time is the magnitude of the final velocity vector, which is the vector sum of the horizontal and vertical velocity components.

To calculate the speed, we use the Pythagorean theorem:

speed = √(horizontal velocity^2 + vertical velocity^2)
= √(6.5 m/s)^2 + (12.48 m/s)^2)
≈ 14.06 m/s

So, the speed at this time is approximately 14.06 m/s.

To determine the direction of travel after the acceleration with respect to the horizontal, we can use trigonometry.

The horizontal component remains unchanged at 6.5 m/s, and the vertical component of velocity is 12.48 m/s. Using the tangent function:

direction = arctan(vertical velocity / horizontal velocity)
= arctan(12.48 m/s / 6.5 m/s)
≈ 61.02°

However, the question asks for the direction to be given between -180° and +180°. To adjust the answer to fit within this range, we can determine the angle's quadrant.

Since the horizontal velocity is positive (to the right) and the vertical velocity is positive (upwards), the direction angle is in the first quadrant. Therefore, the direction of travel after the acceleration with respect to the horizontal is approximately 61.02°.