A particle experiences a constant acceleration that is north at 100m/s^2. At t=0, its velocity vector is 70m/s east. At what time will the magnitude of the velocity be 100m/s?
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The east component will remain
Vx = 70 m/s
The west componen is
Vy = (1/2) a t^2 = 50 t^2
|V|= 100 when Vx^ + Vy^2 = 10^4 m^2/s^2
Write that equation in terms of t and solve for t.
To find the time at which the magnitude of the velocity will be 100m/s, we need to use the kinematic equations of motion and the given information.
The velocity of the particle at any time (t) can be calculated using the equation:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, the initial velocity (u) is 70m/s east. Since the acceleration is north at 100m/s^2, it does not affect the east component of velocity. So, only the magnitude of the velocity changes.
Now, let's rearrange the above equation to solve for time (t):
t = (v - u) / a
In this equation, let's substitute the values:
v = 100m/s (magnitude of the velocity)
u = 70m/s (initial velocity)
a = 0m/s^2 (since acceleration only acts in the north direction)
t = (100m/s - 70m/s) / 0m/s^2
Here we have a problem. Dividing by zero is not defined, and since the acceleration in the east direction is 0, it means that the magnitude of the velocity will never reach 100m/s. The answer to this problem is undefined or infinite.
Therefore, the magnitude of the velocity will never be 100m/s.