A particle's constant acceleration is north at 109 m/s2. At t = 0, its velocity vector is 60.7 m/s east. At what time will the magnitude of the velocity be 114 m/s?
Vy = 109 t
Ve = 60.7
114 = sqrt (109^2t^2 + 60.7^2
12996 = 109^2 t^2 + 60.7^2
To solve this question, we can use the equations of motion to find the time at which the magnitude of the velocity will be 114 m/s.
First, let's break down the given information:
Acceleration (a) = 109 m/s^2 north (in the positive y-direction)
Initial velocity (u) = 60.7 m/s east (in the positive x-direction)
Final velocity (v) = 114 m/s (magnitude, direction not specified)
We can use the equation of motion that relates final velocity (v), initial velocity (u), acceleration (a), and time (t):
v = u + at
Rearranging the equation to solve for time (t):
t = (v - u) / a
Substituting the given values into the equation:
t = (114 m/s - 60.7 m/s) / 109 m/s^2
Simplifying the equation:
t = 53.3 m/s / 109 m/s^2
t = 0.49 s
Therefore, at approximately t = 0.49 seconds, the magnitude of the velocity will be 114 m/s.