Test: Translational Motion and Calculations - 1 - MCAT MCQ

We can use the distance formula d = vt to figure out the problem. The two equations would be x = 45t and x + 600 = 60t.
We can approach the problem more intuitively by looking at the difference between their velocities, which is 15 m/s.
Every second, the police car can make up 15 meters in its chase, and the distance between them is 600 meters.
600 meters divided by 15 m/s gives us a value of 40 seconds.

Average velocity is change in displacement over time, but if the car were traveling at a constant acceleration then the average velocity would be 30 m/s.

Average speed is distance over time. If we were to assume travel in a straight line in one direction only, then they would be the same.

The instantaneous velocity at 1 minute is 20 m/s and at 2 minutes is 40 m/s. In the time interval, velocity could take on any value, positive or negative.

The average acceleration is change in velocity over time. 20 m/s divided by 60 sec is equal to 0.33 m/s² squared.

The force of air resistance is proportional to the square of the velocity. At the beginning, there is no upward force. As the object falls through the sky, the force of air resistance increases and increases to oppose the force due to gravity.

As time progresses further, the force of air resistance will increase until it is equal in magnitude and opposite in direction to the weight.

Its speed does not remain constant for the fall and does not decrease at a constant rate since it gain more speed as it falls.

The speed does increase, but not at a constant rate and not for the entirety of its fall.

Its acceleration, which is gravity, does decrease until the object reaches its terminal velocity, at which point acceleration is zero or velocity is constant.

Instantaneous velocity represents the velocity at a specific instant in time.

By saying the aircraft has a maximum cruising speed of 400 miles per hour, it is still only a potential value to attain. When or whether it actually reaches this velocity is unknown.

Knowing the velocity and time will often aid in figuring out the displacement, given a constant velocity or average velocity, like traveling 15 mph for 5 hours.

Knowing the distance and time will often aid in figuring out an average velocity, like traveling 100 miles in an hour.

If the projectile was traveling upward at 50 m/s at 10 sec after its launch, then we know its instantaneous velocity.

The magnitude of acceleration can be determined by looking at the slope of the tangent at a given point.

At point B, the tangent has a positive slope, and at point C, the tangent has a negative slope. While the slope is negative at point C, the tangent is steeper, so C has a greater magnitude of acceleration or B a lesser magnitude of acceleration.

From the starting point, for the first 3 minutes, the car is traveling away in the positive direction. For a very brief period after 3 minutes, where the graph dips below the x-axis, the car travels back in the negative direction.

Displacement is greater at point C because the car only starts to travel back in the negative direction, which does not make up for the distance traveled after point B. Acceleration is lesser in magnitude at point B, and displacement is greater at point C.

On a velocity-time graph, there is zero acceleration whenever the particle goes from a positive to negative acceleration or vice versa.

Positive acceleration is indicated by tangents with positive slope like point B, and negative acceleration by tangents with negative slope like point C, so between point B and C at the maximum, there is zero acceleration.

Between C and D, there is a global minimum, which indicates zero acceleration.

Look for local minima and maxima, and there is a maximum before A and a minimum at A, so there are 4 instances.

Moving at a constant velocity means there is no acceleration or no unbalanced forces.

Look for a set of forces that are balanced, equal but opposite, so any diagrams with only one force vector can be eliminated:

In the diagram with the forces at 120 to each other, there is an unbalanced force to the left and would slow the object down.

The two forces must be at 180° to each other in order to balance out, but the forces can be at any angle to the velocity vector.

Positive velocity is travel in one direction and negative velocity is travel in the other direction. According to the graph, the direction changes at 4.5 and 6.25 minutes.

The area under the curve is displacement, and from 4 to 5 minutes the area is on both sides of the x-axis, which cancels each other out. The area with the greatest displacement would be from 1.5 to 3.5 minutes.

The slope represents the change in velocity or acceleration. Where the graph is horizontally flat or forms a crest or trough, the slope is zero. From 1.5 to 3.5 minutes, the velocity does not change.

When the graph crosses the x-axis, the velocity is zero. The graph crosses the x-axis at 3 points, so the car has zero instantaneous velocity 3 times.

The slope of the tangent represents velocity, and the greater the absolute value of the slope, the greater the velocity. From 4 to 5 hours, although the slope is negative, the car travels with the greatest velocity.

The car comes to halt when the graph is horizontal or forms a crest or trough. The car stops twice during the trip during 2 to 4 hours and then 5 to 6 hours.

The car has the greatest acceleration when velocity changes the most. Any straight line portion represents constant velocity. The points of greatest acceleration are where the graph forms angles: 2, 4, 5, and 6 hours.

The speed can be found by taking the absolute value of the slope of the tangent. The slope of the tangent at 4.5 hours, while it is negative, is greater than at 1 hour.

The displacement is the shortest distance from the initial to the final position. Since we start and end at the same place, displacement is zero.

The average velocity is calculated with the displacement over time, which would also be zero.

The car could travel at any velocity or speed since we only know it traveled 150 miles in 3 hours. The velocity could have oscillated back and forth from 90 mph to 10 mph or remain constant at 50 mph.

The average speed is calculated with the distance divided by time, which would be 300 miles divided by 8 hours. 300 / 8 = 37.5 mph.