What is the force on a 1000kg elevator that is falling freely at 9.8 m/sec^2
1 answer:
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Answer:
The time that passes until the police catch the speeder is 82.6204 seconds.
Explanation:
A body performs a uniformly accelerated rectilinear motion or uniformly varied rectilinear motion when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases its modulus in a uniform way.
The position is calculated by the expression:
x = x0 + v0*t + 1/2*a*t²
where:
- x0 is the initial position.
- v0 is the initial velocity.
- a is the acceleration.
- t is the time interval in which the motion is studied.
First, let’s look at the police car’s equations of motion. In this case:
- x0= 0
- v0= 50 m/s
- a= 2 m/s²
So: x = 50 m/s*t + 1/2*2 m/s²*t²
Now for the speeder’s car’s equations of motion you know:
- x0= 3 km= 3,000 m
- v0= 55 m/s
- a= 1 m/s²
So: x = 3,000 m + 55 m/s*t + 1/2*1 m/s²*t²
When the police catch the speeder they are both in the same position. So:
50 m/s*t + 1/2*2 m/s²*t²= 3,000 m + 55 m/s*t + 1/2*1 m/s²*t²
Solving:
0= 3,000 m + 55 m/s*t + 1/2*1 m/s²*t² - 50 m/s*t - 1/2*2 m/s²*t²
0= 3,000 + 55 *t + 1/2*t² - 50*t - 1*t²
0= 3,000 + 55 *t - 50*t - 1*t² + 1/2*t²
0= 3,000 + 5*t - 1/2*t²
Applying the quadratic formula:
x1= -72.6209
and x2= 82.6209
Since you are calculating the value of a time and it cannot be negative, then <u><em>the time that passes until the police catch the speeder is 82.6204 seconds.</em></u>