Solving Motion in a Plane: Find Velocity at t=7.6s

[tag16]

A particle experiences a constant acceleration that is south at 2.3 m/s2. At t = 0, its velocity is 43.4 m/s east. What is its velocity at t = 7.6 s?

I got 60.8 m/s but it also asks for the direction in degrees and I'm not sure how to find that.

 

[tiny-tim]

Welcome to PF!

Hi tag16! Welcome to PF!

I got 60.8 m/s but it also asks for the direction in degrees and I'm not sure how to find that.

Dividing the y by the x will give you the tangent of the angle

(and remember that the angle increases anticlockwise! )

 

[nlightner]

Based on the information provided, the particle's initial velocity is 43.4 m/s east and it experiences a constant acceleration of 2.3 m/s2 south. To find the velocity at t=7.6s, we can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Plugging in the values, we get:

v = 43.4 m/s east + (2.3 m/s2 south)(7.6 s)

v = 43.4 m/s east + 17.48 m/s south

Using the Pythagorean theorem, we can find the magnitude of the velocity:

|v| = √(43.4^2 + 17.48^2) = 46.57 m/s

To find the direction in degrees, we can use the inverse tangent function:

θ = tan^-1(17.48/43.4) = 22.85 degrees south of east

Therefore, the particle's velocity at t=7.6s is 46.57 m/s at an angle of 22.85 degrees south of east.