(a) 18717 N
Newton's second law in this situation can be written as:
(1)
where
T is the tension in the cable, pointing upward
W is the weight of the elevator+passengers, pointing downward
m is the mass of the elevator+passengers (1700 kg)
a is the acceleration of the system (1.20 m/s^2, upward)
The weight is equal to the product between the mass, m, and the gravitational acceleration, g:
So now we can solve eq.(1) to find T, the tension in the cable:
(b) 16677 N
In this situation, the elevator is moving with constant velocity: this means that its acceleration is zero,
a = 0
So Newton's second law becomes
and so we find
(c) 15657 N
During the deceleration phase, Newton's second law can be written as:
(1)
Where the acceleration here points downward (because the elevator is decelerating), as the weight W, so we can write it as a negative number:
a = -0.600 m/s^2
we can solve the equation to find T, the tension in the cable:
(d) 19.35 m, 0 m/s
Distance covered during the first part of the motion; we know that
u = 0 is the initial velocity
a = 1.20 m/s^2 is the acceleration
t = 1.50 s is the time
So the distance covered is given by
and the final velocity after this phase is
During the 2nd part of the motion, the elevator moves at constant speed of 1.8 m/s for t=8.50 s, so the distance covered here is
Finally, in the third part the elevator decelerates at a = -0.600 m/s^2 for t = 3.00 s. So, the distance covered here is
and the final velocity is
and the total distance covered is
Answer:
Same direction: t=234s; d=6.175Km
Opposite direction: t=27.53s; d=0.73Km
Explanation:
If the automobile and the train are traveling in the same direction, then the automobile speed relative to the train will be (<em>the train must see the car advancing at a lower speed</em>), where
is the speed of the automobile and
the speed of the train.
So we have .
So the train (<em>anyone in fact</em>) will watch the automobile trying to cover the lenght of the train L at that relative speed. The time required to do this will be:
And in that time the car would have traveled (<em>relative to the ground</em>):
If they are traveling in opposite directions, <u>we have to do all the same</u> but using (<em>the train must see the car advancing at a faster speed</em>), so repeating the process: