Answer:
acceleration
Explanation:
acceleration =velocity final-velocity initial /time
Answer:
x = 1474.9 [m]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of forces must be equal to the product of mass by acceleration.
We must understand that when forces are applied on the body, they tend to slow the body down to stop it.
So as the body continues to move to the left, it is slowing down. Therefore we must calculate this deceleration value using Newton's second law. We must perform a sum of forces on the x-axis equal to the product of mass by acceleration. With leftward movement as negative and rightward forces as positive.
ΣF = m*a
![10 +12*sin(60)= - 6*a\\a = - 3.39[m/s^{2}]](https://tex.z-dn.net/?f=10%20%2B12%2Asin%2860%29%3D%20-%206%2Aa%5C%5Ca%20%3D%20-%203.39%5Bm%2Fs%5E%7B2%7D%5D)
Now using the following equation of kinematics, we can calculate the distance of the block, before stopping completely. The initial speed must be 100 [m/s].

where:
Vf = final velocity = 0 (the block stops)
Vo = initial velocity = 100 [m/s]
a = - 3.39 [m/s²]
x = displacement [m]
![0 = 100^{2}-2*3.39*x\\x=\frac{10000}{2*3.39}\\x=1474.9[m]](https://tex.z-dn.net/?f=0%20%3D%20100%5E%7B2%7D-2%2A3.39%2Ax%5C%5Cx%3D%5Cfrac%7B10000%7D%7B2%2A3.39%7D%5C%5Cx%3D1474.9%5Bm%5D)
Answer:
If a body is constantly accelerating the graph would be one-quadrant
The trains take <u>57.4 s</u> to pass each other.
Two trains A and B move towards each other. Let A move along the positive x axis and B along the negative x axis.
therefore,

The relative velocity of the train A with respect to B is given by,

If the train B is assumed to be at rest, the train A would appear to move towards it with a speed of 170 km/h.
The trains are a distance d = 2.71 km apart.
Since speed is the distance traveled per unit time, the time taken by the trains to cross each other is given by,

Substitute 2.71 km for d and 170 km/h for 

Express the time in seconds.

Thus, the trains cross each other in <u>57.4 s</u>.