General Principles - MCAT Physical

Choice 3 is correct. Recall that physics problems tell you where to begin and end. The only thing that matters in figuring displacement is the beginning point and end point. The man effectively walked along the hypotenuse of a right triangle whose sides measured 2 km each. Since a2 + b2 = c2, then 4 + 4 = c2, and the correct response is the square root of 8 with the direction of the displacement added for clarity.

Scalar quantities are those that can be described with magnitude only, as opposed to vectors, which include both magnitude and direction components. Distance, speed, and time are all scalars. Displacement is not a scalar, as it involves both the distance and the direction moved from a starting point. Velocity also includes a direction component, and is therefore a vector quantity.

Choice 4 is correct. In physics, it is often helpful to draw a diagram of the problem. In this case, the respondent would draw a vector 10 hours x 10 knots from south to north (usually going towards the top of a sheet of paper) and a perpendicular vector 50 nautical miles long projecting from the arrowhead of the first vector to the left, or west. The hypotenuse of the right triangle so created is the actual path of a boat experiencing these forces. Without resorting to calculating the square root of the sum of (10,000 plus 2,500 nautical miles2), it is obvious that choice 4 is the only reasonable one.

To find total displacement, find the vector sum of the northern and eastern travel using the Pythagorean theorem.

Note that displacement differs from total distance traveled (which would be 650km).

Since it takes the body twice the time to cover the same distance, the magnitude of v2 is half of v1. Also note that the direction of the body has reversed (velocity is a vector quantity), and so v2 must be negative, with regard to v1.

v = d/t

v1 = d/2

v2 = -d/4

v2 = (-1/2)v1

It is stated that between seconds 3 and 6 the object's acceleration increases. On a velocity vs. time graph, this would translate into a nonlinear curve.

Between seconds 1 and 3 the acceleration is constant, and would result in a linear velocity-time relationship. When the object travels at a constant velocity, the acceleration, which correlates to the slope, is zero.

Acceleration is equal to the acceleration due to gravity (about 10 m/s2) and is constant while the projectile is in the air. The force on the object is given by the product of its mass and gravitational acceleration, and is also constant.

Since the particle is decelerating the acceleration will be in the opposite direction of the movement. The particle is moving to the right, therefore its acceleration must be to the left.

When the rope is around the pulley, the acceleration is slower due to internal friction, thus the initial acceleration of the platform would be lower. Once the platform was in free fall, however, the acceleration due to gravity is constant at 9.8 m/s2; thus, the final acceleration of the platform is the same in both scenarios.

The three blocks must remain in contact as they move, so they will each have the same velocity and acceleration regardless of their different masses. So, .

To solve this question, you must understand the difference between total distance and total displacement. While it is true that the baseball traveled a specific distance, the displacement is the final position of the baseball relative to the original position of the ball. The ball started and ended in the same place, so its total displacement is 0m.

At its peak height, the ball will have a velocity of ; however, the acceleration on the ball due to gravity is constantly experienced by the baseball. As a result, while in the air, the ball will always have a vertical acceleration of

This is a question that helps you see how displacement and velocity compare to distance and speed, respectively. Remember that the runner starts and stops at the same point on the track, so her total displacement is zero feet.

Since , the runner's average velocity is . Since the runner's average speed is greater than her average velocity, we can conclude this is the correct answer. Her average speed refers to her distance, rather than displacement, and is greater than zero.

This is also important to remember! Since the runner changed direction on the track, she accelerated and decelerated accordingly based on her position on the track. She did not have a constant acceleration.

Since we need to solve for the total distance around the track.

Now, we can solve for speed using our equation. With 60 seconds per minute, 3 minutes is equal to 180 seconds.

Keep in mind that the question asked for the average speed, not velocity. Had it asked for the average velocity, would be the correct answer, as the total displacement is zero.

This question requires us to find the end position relative to the starting position. As a result, 150m is not the correct answer; this would be the distance, not the displacement.

The first portion involves writing down that the person moves north 100m. At this point in time, his distance on the y-axis is +100m. Next, we have to see how the person's x-axis and y-axis distances change after turning left 120o, and walking 50m.

For the x-axis, we use the equation based on a right triangle. By turning 120o, the man is walking at an angle of 30o to the x-axis. Since this is the only movement in the x-axis, the total distance on the x-axis is –43.3m (movement to the left will be negative).

For the y-axis, we use the equation Since this 25m is heading downward, we subtract it from the initial upward movement of 100m. As a result, the total movement in the y-axis is 75m.

Now that we have the component vectors, we can use the pythagorean theorem to determine the total displacement of the man.

Keep in mind that velocity is a vector, and a change in direction results in a change in velocity. When a runner runs in a circular manner around a track, the velocity of the runner will oscillate between a maximum positive velocity and a minimum negative velocity. As a result, this scenario matches the velocity vs. time graph appropriately.

A very quick and easy way to determine the total displacement of an object is to graph the velocity of the object over time, and calculate the area under the line. The velocity vs. time graph would show a triangle with a height of and a base of 15 seconds. The area of a triangle is determined by the equation .

As a result, we conclude that the sprinter has a total displacement of 75m.

At maximum height, the projectile is still moving in the x-direction. The horizontal component of velocity remains constant during projectile motion, therefore .

Gravity is always acting on any body in projectile motion, therefore . Acceleration will be constant at .

Inconstant velocity implies a non-zero acceleration. A non-zero acceleration implies a non-zero force.

Circular motion is made possible by acceleration along the radius of the circle. Since there is acceleration, the linear velocity is not constant even if angular velocity is constant.

The other scenarios describe instances in which there is no net force, and thus no net acceleration or change in velocity.

Displacement is a vector quantity, so we cannot simply add the distances. Instead, we will need to triangulate the position of the plane.

We can draw the plane's path on a set of coordinate axes. The initial displacement will be in the vertical direction. The plane will then travel an equivalent distance at an angle of to the horizontal ( to the original flight path), essentially with a slope of one. By bisecting the angle created by the two flight paths, we can create two similar right triangles with hypotenuse lengths of . The measure of the bisected angle will be .

We need to solve for the side opposite the bisected angle, given the length of the hypotenuse. We will use the sine function.

This is the length on only one of the triangle sides. We need to double this distance to find the total displacement.