A particle moving at a velocity of 4.8 m/s in the positive x direction is given an acceleration of 3.4 m/s2 in the positive y direction for
4.4 s. What is the final speed of the particle? Answer in units of m/s.
V = Vo + a*t(i) = 4.8 + 3.4i*4.4 = 4.8 + 14.96i = sqrt(4.8^2 + 14.960^2) = 15.71 m/s.
To find the final speed of the particle, we need to use the equations of motion.
First, let's find the final velocity in the y-direction using the equation:
vf = vi + at
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.
Given that the initial velocity in the y-direction is 0 m/s (since the particle is originally moving only in the positive x-direction), the equation becomes:
vf_y = 0 + (3.4 m/s^2)(4.4 s)
vf_y = 14.96 m/s
Next, we can find the final velocity in the x-direction using the equation:
vf_x = vi_x
where vf_x is the final velocity in the x-direction and vi_x is the initial velocity in the x-direction.
The initial velocity in the x-direction is given as 4.8 m/s, so the final velocity in the x-direction is also 4.8 m/s.
Now, we can use the Pythagorean theorem to find the magnitude of the final velocity (Vf) using the equation:
Vf = sqrt((vf_x)^2 + (vf_y)^2)
Plugging in the values, we have:
Vf = sqrt((4.8 m/s)^2 + (14.96 m/s)^2)
Vf = sqrt(23.04 m^2/s^2 + 223.8016 m^2/s^2)
Vf = sqrt(246.8416 m^2/s^2)
Vf ≈ 15.71 m/s
Therefore, the final speed of the particle is approximately 15.71 m/s.