an object moves in the -x direction at a speed of 30 m/s. As it passes through the origin it starts to experience a constant acceleration of 1.4 m/s^2 in the +x direction. How much time elapses before it returns to the origin?
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so when is velocity zero?
Vf=Vi+at
0=-30+1.4t
t= 30/1.4 seconds
To find the time it takes for the object to return to the origin, we can use the concept of relative velocity and the equations of motion.
Let's denote the initial velocity of the object as u = -30 m/s (negative because it is moving in the -x direction) and the acceleration as a = 1.4 m/s^2.
The first step is to determine when the object comes to a stop and changes its direction from -x to +x. We can find this by using the equation of motion:
v = u + at
where v is the final velocity, t is the time, and the object's velocity changes from -x to +x, so v = 0.
0 = -30 m/s + (1.4 m/s^2)t
Solving this equation, we find:
30 m/s = 1.4 m/s^2 * t
t = 30 m/s / 1.4 m/s^2
t ≈ 21.43 s
So, it takes approximately 21.43 seconds for the object to come to a stop and change direction.
Next, we need to find the time it takes for the object to travel back to the origin, considering only the positive acceleration. We can again use the equations of motion, but this time the initial velocity will be 0 m/s and the acceleration will be 1.4 m/s^2.
We can use the equation:
s = ut + (1/2)at^2
where s represents the displacement. In this case, since the object is returning to the origin, the displacement will be zero (s = 0).
0 = 0 + (1/2) * (1.4 m/s^2) * t^2
Solving this equation:
0 = 0.7 m/s^2 * t^2
As the equation implies, the only solution is when t = 0.
Therefore, the object will return to the origin exactly at t = 21.43 seconds.
To summarize, the time that elapses before the object returns to the origin is approximately 21.43 seconds.