a particle moving at a velocity of 6 m/s in the positive x direction is given an acceleration of 3.9m/s^2 in the positive Y direction for 3.4s. what is the final speed of the particle? answer in units of m/s
The x component of velocity remains
Vx = 6 m/s.
The y component of velocity increases with time, t, and is
Vy = 3.9 t. After 3.4 seconds
Vy(@ t=3.4s) = 13.3 m/s
The final speed (the magnitude of velocity) is sqrt(Vx^2 + Vy^2)
which is about 14.6 m/s
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heyeyere
To find the final speed of the particle, we need to combine the initial velocity and the acceleration to determine the change in velocity. Then, we can add this change in velocity to the initial velocity to get the final speed.
Given:
Initial velocity (ui) = 6 m/s
Acceleration (a) = 3.9 m/s^2
Duration (t) = 3.4 seconds
Since the acceleration is in the positive y direction, it does not affect the initial velocity in the positive x direction. Therefore, the change in velocity in the x-direction is 0 m/s.
The change in velocity in the y-direction (delta vy) can be calculated using the equation:
delta vy = a * t
Substituting the given values, we get:
delta vy = 3.9 m/s^2 * 3.4 s
delta vy = 13.26 m/s
Now, we can calculate the final speed (v) using the Pythagorean theorem:
v = sqrt((ui + delta vx)^2 + (delta vy)^2)
In this case, delta vx is 0 m/s, so we have:
v = sqrt((6 m/s)^2 + (13.26 m/s)^2)
v = sqrt(36 m^2/s^2 + 175.2276 m^2/s^2)
v = sqrt(211.2276 m^2/s^2)
v ≈ 14.53 m/s
Therefore, the final speed of the particle is approximately 14.53 m/s.