Contact Forces Problem

A crate resting on a floor has weight, normal, friction, and Force forces.  The normal and weight forces are balanced.  The friction and Force forces aren't balanced because the object is moving forward therefore there is a positive force to the right.  Friction is negative because it takes away force by slowing it down.

To find the weight force, mass multiplied by acceleration is enforced (ma=W).  The weight force is negative because acceleration is negative (-9.8m/s2).

Normal force and weight force are balanced, so the normal will equal the weight force except a positive number.  The normal and weight forces added together should equal zero.

Next, the frictional force should be calculated using the friction equals mu multiplied by Normal equation. N signifies normal force, while M represents the coefficient of friction.  The coefficient of friction is given (0.15) and is mulitplied by the Normal just calculated (490N) to find friction (73.5N).

Finally to calculate acceleration, the equation Net Force equals mass multiplied by acceleration should be used.  EF signifies net force which will be Force subtracted by friction.  The mass is given in the problem (50kg).  To solve for acceleration, subtract the friction force from the Force.  Then, divide that number by the mass to find acceleration.

Acceleration Problem

Since the displacement is less than 13 meters, the driver will not hit the child given a reaction time of 0.25 seconds.

Conversions:
Since the problem gives us the speed limit in 40km divided by one hour times 1 hour over 3600 seconds times 1000 meters over 1 kilometer gives us 11.1 m/s.

xt graph shape:
The problem states the drivers reaction time is 0.25 seconds.  Therefore, the driver is moving at constant speed for the first 0.25 seconds of the graph.  At 0.25 seconds, the driver reacts to the child in the street and begins to slow his car by breaking.  Since time is still moving forward and also the car is moving forward, the graph has a y squared equals mx plus b equation.

vt graph shape:
In the first 0.25 seconds of the graph, the car is moving at 11.1 meters per second forward.  Since the car is moving at a constant velocity forward, this correlates into a horizontal line on the vt graph.  At 0.25 seconds, the driver reacts and begins to break.  His acceleration is -8 meters per a second squared.  On the graph, there is a downward linear line to reciprocate the driver breaking until stopping.

at graph shape:
In the first 0.25 seconds, the car is moving forward at a constant velocity.  Since the car is not accelerating at this moment, the graph begins at t equals zero for 0.25 seconds.  After 0.25 seconds, the car begins to negatively accelerate.  This is represented on the graph by a horizontal line in the negative quadrant.  The line is at -8 meters per second squared, because that is the drivers acceleration.  The driver finally comes to a stop represented by the time returning to t equals zero.

A equals change in velocity over change in time explanation:
The problem wants to figure out if the automobile is able to stop before hitting the child.  The child is at 13 meters.  Since the acceleration of the vehicle and the initial velocity are given in the problem, I can use these to find the amount of time the vehicle takes to stop.  The vehicle stops after 1.387 seconds after applying the brake.

A equals lw+1/2bh:
After I find the change in time the driver took, I am able to find the displacement under the line of the vt graph to find the displacement of the automobile.  I can further divide the area up into a rectangle and a triangle.  Using the formulas length times width for a rectangle and 1/2 base times height for a triangle, I will be able to find the area which is the displacement by adding them together.

lw (the rectangle) explanation:
The length of the rectangle in the vt graph is the intial velocity the vehicle is going at.  The width is the amount of time (reaction time) 0.25 seconds the driver takes to realize to begin to brake.  The area of the rectangle is the displacement of the reaction time of the driver.

1/2bh explanation:
The driver begins to brake forming a triangle on the vt graph of the second half of the graph.  I found the change in time earlier with acceleration equals change in velocity over change in time of the vehicle.  I multiply all these together times 1/2 to find the displacement of the vehicle.

Solution and Conclution:
After adding the areas of the triangle and rectangle under the velocity time graph line, the distance the car went is produced.  The car went 10.5 meters away from the car.  The child was safe from the car by 2.5 meters.

Projectile Motion Problem

The y-graph represents the problem, because the car begins at above the origen then is thrown to the Earth after a certain amount of time goes by.  The slope is always negative, so the derivative and Vy graph is a negative linear line.  As time goes by, the position approaches 0.  As it approaches 0, it approaches 0 faster.  The object starts at rest and begins going faster and faster in a negative direction.

The x-graph represents the problem, because as time goes on the car is still horizontally moving.  The derivative of a positive linear line the poitive position go in a positive direction at a positive rate.

Equations:
First, I used the position equation.  The car travels from 0 meters to 125 meters, so the position is -125 meters.  The vertical motion doesn't have an initial velocity.  Gravity pulls on the car at -9.8 m/s2.  I then solve the equation to find the time the car took to reach the ground.  I find after 5.05 seconds, the tornado replaces the car on the ground.

Next, I use the change in x equation, change in x equals Vx times t, to find how far horizontally the car actually travels.  The tornado flings the car horizontally at an initial velocity of 90.0m/s.  Since I already found the time the car takes to reach the ground, I can plug 5.05 seconds in for time.  I come up with the horizontal displacement of the car by the tornado to be 454.5 meters.