A 1000 kg elevator is accelerated upward at a rate of 0.70 m/s2. What is the tension in the cable pulling the elevator upward wh
2 answers:
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Answer:
244.64m
Explanation:
First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:
After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:
1. The velocity will start to decrease untill it reaches 0m/s.
2. Then, the velocity will start to increase at the rate of the acceleration.
The distance that the ball travels in the first phase can be found with the following expression:
Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:
Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:
Then, the time of the second phase will be:
Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:
V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:
So, the total distance covered by this object in meters will be the sum of all the distances we found: