Displacement, Velocity, and Acceleration - MCAT Chemical and Physical Foundations of Biological Systems
Card 1 of 154
A boat sails straight north at 10 knots (10 nautical miles per hour) for ten hours. As it sails, a steady wind comes from the east at 5 knots. If the helmsman makes no correction for the wind, where will the boat be compared to its starting point?
- 50 nautical miles north of the starting point
- 150 nautical miles north of the starting point
- 112 nautical miles northeast of the starting point
- 112 nautical miles northwest of the starting point
- none of the above
A boat sails straight north at 10 knots (10 nautical miles per hour) for ten hours. As it sails, a steady wind comes from the east at 5 knots. If the helmsman makes no correction for the wind, where will the boat be compared to its starting point?
- 50 nautical miles north of the starting point
- 150 nautical miles north of the starting point
- 112 nautical miles northeast of the starting point
- 112 nautical miles northwest of the starting point
- none of the above
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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.
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.
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A body travels 150km north, and then 500km east. Its total displacement is .
A body travels 150km north, and then 500km east. Its total displacement is .
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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).
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).
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A body moves from position X at a constant velocity, v1, for two seconds, remains at rest for three seconds, and then returns to the original position X at a constant velocity v2 after four seconds. Which of the following represents the velocity of v2?
A body moves from position X at a constant velocity, v1, for two seconds, remains at rest for three seconds, and then returns to the original position X at a constant velocity v2 after four seconds. Which of the following represents the velocity of v2?
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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
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
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A graph is made of the following motion by plotting time on the x-axis and velocity on the y-axis. An object accelerates uniformly between for three seconds, and then steadily increases the acceleration for the next three seconds. It remains at a constant velocity for the next two seconds. Which of the following is not true?
A graph is made of the following motion by plotting time on the x-axis and velocity on the y-axis. An object accelerates uniformly between for three seconds, and then steadily increases the acceleration for the next three seconds. It remains at a constant velocity for the next two seconds. Which of the following is not true?
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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.
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.
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For a projectile moving in a parabolic path on Earth, which of the following is true? (Ignore wind resistance.)
For a projectile moving in a parabolic path on Earth, which of the following is true? (Ignore wind resistance.)
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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.
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.
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A particle is traveling to the right with decreasing speed. Which of the following shows the direction of acceleration?
A particle is traveling to the right with decreasing speed. Which of the following shows the direction of acceleration?
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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.
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.
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A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.
If we assumed the pulley had internal friction, how would the initial and final accelerations of the platform with the individual masses change?
A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.
If we assumed the pulley had internal friction, how would the initial and final accelerations of the platform with the individual masses change?
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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.
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.
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Three blocks (in left-to-right order: block A, block B, block C) are pushed along a frictionless horizontal surface by a 50N force to the right, which acts on block A.
If the masses of the blocks are given by 
, 
, and 
, which of the following best describes the relationship between the accelerations of the blocks?
Three blocks (in left-to-right order: block A, block B, block C) are pushed along a frictionless horizontal surface by a 50N force to the right, which acts on block A.
If the masses of the blocks are given by
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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, 
.
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,
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A child throws a ball straight up into the air. He throws the ball with an initial velocity of 
. Assume there is no air resistance.
The child catches the ball in his glove after it falls. What is the total displacement of the ball?
A child throws a ball straight up into the air. He throws the ball with an initial velocity of
The child catches the ball in his glove after it falls. What is the total displacement of the ball?
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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.
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.
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A child throws a ball straight up into the air. He throws the ball with an initial velocity of 
. Assume there is no air resistance.
What is the vertical acceleration of the ball when it is at its peak height?
A child throws a ball straight up into the air. He throws the ball with an initial velocity of
What is the vertical acceleration of the ball when it is at its peak height?
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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 
At its peak height, the ball will have a velocity of
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A runner completes a lap on a circular track in three minutes. The track has a radius of 210ft. The runner ran at a constant speed around the track.
Which of the following statements is true?
A runner completes a lap on a circular track in three minutes. The track has a radius of 210ft. The runner ran at a constant speed around the track.
Which of the following statements is true?
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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.
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
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.
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A runner completes a lap on a circular track in three minutes. The track has a radius of 210ft. The runner ran at a constant speed around the track.
What was the runner's average speed around the track?
A runner completes a lap on a circular track in three minutes. The track has a radius of 210ft. The runner ran at a constant speed around the track.
What was the runner's average speed around the track?
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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.
Since
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,
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A man is walking through a forest. He walks north for 100m. He then turns left at an angle of 120o and walks for another 50m.
What is the man's total displacement?
A man is walking through a forest. He walks north for 100m. He then turns left at an angle of 120o and walks for another 50m.
What is the man's total displacement?
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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.
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.
For the y-axis, we use the equation
Now that we have the component vectors, we can use the pythagorean theorem to determine the total displacement of the man.
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A velocity vs. time graph that shows the plotted line oscillating in a sinusoidal fashion. Which of the following scenarios could be represented by the sinusoidal line on the graph?
A velocity vs. time graph that shows the plotted line oscillating in a sinusoidal fashion. Which of the following scenarios could be represented by the sinusoidal line on the graph?
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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.
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.
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A sprinter positions himself at the starting line. A gun fires, and he begins to accelerate at a constant rate. He reaches his top speed of 
after five seconds, and then decreases his velocity at a constant rate. He reaches a full stop after fifteen seconds of running.
What is the sprinter's total displacement after the fifteen seconds of running?
A sprinter positions himself at the starting line. A gun fires, and he begins to accelerate at a constant rate. He reaches his top speed of
What is the sprinter's total displacement after the fifteen seconds of running?
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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.
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
As a result, we conclude that the sprinter has a total displacement of 75m.
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A projectile is fired using a catapult. It travels in an arc until it hits the ground several meters away. At its maximum height, which of the following is true?
A projectile is fired using a catapult. It travels in an arc until it hits the ground several meters away. At its maximum height, which of the following is true?
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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 
.
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
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Which of the following is an example of inconstant velocity?
Which of the following is an example of inconstant velocity?
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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.
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.
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A plane flies 
North and then flies 
Northeast. What is the magnitude of the plane's final displacement from its original location?
A plane flies
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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.
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
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.
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A car travels at 30mph due east for thirty minutes, then turns left and travels due north at 60mph for ten minutes. Finally, it turns right and travels east again at 60mph for twenty minutes. What are the magnitude and direction of the car's total displacement?
A car travels at 30mph due east for thirty minutes, then turns left and travels due north at 60mph for ten minutes. Finally, it turns right and travels east again at 60mph for twenty minutes. What are the magnitude and direction of the car's total displacement?
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Use the equation 
to find the distance traveled in each part of the trip.
Part 1 (east): 
Part 2 (north): 
Part 3 (east): 
Find the total distance traveled in each direction.
Treat these components as sides of a right triangle, in which the hypotenuse represents the total displacement. Find the hypotenuse of this triangle using the Pythagorean theorem.
Now that we have the magnitude of the displacement, we still need to find the directional angle. The angle north of east can be found with an inverse trigonometry function, such as inverse tangent. We know that we are working with a triangle with sides of north and east, and the hypotenuse equal to the displacement. Orienting the angle, we can see that the opposite side is the northwards displacement, and the adjacent side is the eastwards displacement.
Use the equation
Part 1 (east):
Part 2 (north):
Part 3 (east):
Find the total distance traveled in each direction.
Treat these components as sides of a right triangle, in which the hypotenuse represents the total displacement. Find the hypotenuse of this triangle using the Pythagorean theorem.
Now that we have the magnitude of the displacement, we still need to find the directional angle. The angle north of east can be found with an inverse trigonometry function, such as inverse tangent. We know that we are working with a triangle with sides of north and east, and the hypotenuse equal to the displacement. Orienting the angle, we can see that the opposite side is the northwards displacement, and the adjacent side is the eastwards displacement.
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If a car is traveling around a circular track and returns to its starting point every two minutes, what is its average velocity?
If a car is traveling around a circular track and returns to its starting point every two minutes, what is its average velocity?
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The average velocity is equal to displacement over time.
The displacement of the race car is zero because the ending position was the same as the starting position. The average velocity will therefore also be equal to zero.
Remember that velocity is a vector, while speed is a scalar. Velocity is measured in terms of displacement and speed is measured in terms of distance.
The average velocity is equal to displacement over time.
The displacement of the race car is zero because the ending position was the same as the starting position. The average velocity will therefore also be equal to zero.
Remember that velocity is a vector, while speed is a scalar. Velocity is measured in terms of displacement and speed is measured in terms of distance.
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