All High School Physics Resources
Example Questions
Example Question #3 : Freefall
A tennis ball is thrown straight up and it is caught at the same height the person released the ball from their hand. Which of the following is false? Ignore air resistance.
The velocity will change sign at the the top of the motion
The time the tennis ball is traveling up is equal to the time it falls down
The speed of the tennis ball as it leaves the person's hand is the same as when it is caught
All of these answers are true
Acceleration and velocity point in the same direction the entire time
Acceleration and velocity point in the same direction the entire time
The question asks to point out the false statement. Everything on Earth is accelerated downwards by gravity, all the time, by . Think of gravity as having a negative sign. When the ball is thrown up, acceleration is working against the velocity slowing the ball down. Their signs are opposite. But when the ball is falling back down to Earth, the velocity and acceleration have the same sign. So velocity and acceleration will not always have the same sign.
If gravity could be said to have a negative sign since it pulls everything downward, then an upward velocity would have a positive (and opposing sign). At the top of the trajectory when the ball's upward velocity is finally overcome by gravity, the sign of the velocity becomes negative as it now points back down to Earth. So it is true velocity changes sign at the top.
Since the ball is caught at the same height, both the time up and time down are equal and it will be traveling at the same speed. Since gravity is the only force acting on it, the ball loses all its velocity on the way up and regains that exact amount by the time it reaches the height it started the journey. This also makes the time up equal to the time it falls. If the same force acts with same strength (gravity) the entire time, why would either of these change?
Example Question #4 : Freefall
A rocket rises vertically from rest, with an acceleration of until it runs out of fuel at an altitude of . After this point, its acceleration is that of gravity downward. How long (total) is it in the air from the time it is launched to when it lands back on the ground?
This is a multi-step problem and involves looking at the rocket at several points.
Point 0 - launch from the ground
Point 1 - when the fuel runs out
Point 2 - the height that the rocket reaches
Point 3 - when the rocket reaches the ground
Therefore all knowns and unknowns will be denoted by these different points.
To find the total time, it will be necessary to determine the time traveled under the acceleration, the time time traveled without acceleration to the peak and the time from the top of the peak to the ground.
Point 0 to Point 1
Equations:
The initial velocity is zero which simplifies the equation to
Rearrange to solve for time
It will also be important to know the speed that the rocket is traveling at when it is done accelerating.
Point 1 to Point 2
Equations:
Rearrange to solve for time
This time is the total time it takes to accelerate then reach the highest point of the peak.
It will also be important to know how high the rocket goes.
This is the total height that it takes the rocket to reach the highest point of the peak.
Point 2 to Point 3
Equations:
The initial velocity is zero which simplifies the equation to
Rearrange to solve for time
This is the total time for the rocket to go up and then come back down.
Example Question #1 : Freefall
Leslie rolls a ball out of a window from above the ground, such that the initial y-velocity is zero. How long will it be before the ball hits the ground?
We are given the initial velocity, acceleration, and distance traveled. Using the equation below, we can solve for the time. Remember that the initial velocity is 0m/s so it drops out of the equation.
The distance is negative, which makes since because the ball is traveling downward. Also when taking the square root, only the positive value is needed as it is impossible to have negative time.
Example Question #1 : Freefall
A ball is thrown vertically with a velocity of . What is its velocity at the highest point in the throw?
There is insufficient information to solve
When examining vertical motion, the vertical velocity will always be zero at the highest point. At this point, the acceleration from gravity is working to change the motion of the ball from positive (upward) to negative (downward). This change is represented by the x-axis on a velocity versus time graph. As the ball changes direction, its velocity crosses the x-axis, momentarily becoming zero.
Example Question #1 : Representations Of Motion
Calculate the total displacement of the object.
This is a velocity versus time graph. The displacement can be determined from the area under the curve of the graph.
Begin with the first . This is a triangle with a base of and height of .
Next look at the next . This is a triangle with a base of and height of . That negative is important as it indicates that the object was traveling in the opposite direction.
For the final , the area is that of a rectangle with a base of and a height of . Again this negative is important as it indicates what direction the object was traveling in.
To find the total displacement, add all of the areas together.
Therefore the final displacement is .
Example Question #2 : Representations Of Motion
What is the object’s total distance traveled at ? What is the object’s displacement at ?
This is a position versus time graph. The total distance traveled is the total position that has been traveled during the entire time.
From 0 to 10 seconds, the object traveled .
From 10 seconds to 15 seconds, the object did not move.
From 15 seconds to 30 seconds the object traveled backward.
The distance is the total amount traveled.
The displacement which is the distance traveled from the original position. Since at the object has returned to the , the displacement is .
Example Question #3 : Representations Of Motion
What is the acceleration of the object at ?
This is a position versus time graph. To find the velocity of an object at a particular time, determine the slope of the graph at that time. Using the points and we can determine the slope of the graph at 5 seconds since it is a constant slope.
Example Question #4 : Representations Of Motion
You are given a graph of displacement vs. time. Which of the following ways can be used to determine velocity at any given point?
The velocity is the y-intercept of the graph
The velocity is the total change in displacement over the total time
The velocity cannot be determined from a graph of displacement vs time
The velocity is the area under the curve for any time interval
The velocity is the slope of the line at any given time
The velocity is the slope of the line at any given time
The important thing to note is that the question asks for the velocity at any given point. The average velocity will be equal to the total displacement divided by the total time, but the question is asking for the instantaneous velocity.
Velocity is calculated by a change in displacement over a change in time. In a displacement versus time graph, this is equal to the slope.
This tells us that the slope at a certain time will be equal to the velocity at that time.
In calculus terms, velocity is the derivative of the function for displacement in terms of time. What this means is that at any point on the graph, the instantaneous slope is the velocity for that given time. To determine velocity, one must find the slope of the line at that particular time interval.
Example Question #5 : Representations Of Motion
You are given a graph of velocity vs. time. Which of the following ways can be used to determine acceleration at any given point?
The acceleration is the total change in velocity over the total time
The acceleration is the slope at any particular point in time
The acceleration is the area under the curve for any given time interval
The acceleration is the y-intercept of the velocity vs time graph
The acceleration cannot be determined from the graph
The acceleration is the slope at any particular point in time
The important thing to note is that the question asks for the acceleration at any given point. The average acceleration will be equal to the net change in velocity divided by the total time, but the question is asking for the instantaneous acceleration.
Acceleration is calculated by a change in velocity over a change in time. In a velocity versus time graph, this is equal to the slope.
This tells us that the slope at a certain time will be equal to the acceleration at that time.
In calculus terms, acceleration is the derivative of the function for velocity in terms of time. What this means is that at any point on the graph, the instantaneous slope is the acceleration for that given time. To determine acceleration, one must find the slope of the line at that particular time interval.
Example Question #6 : Representations Of Motion
Describe the motion of the object.
The object is slowing down for 10 seconds. Then the object travels at a constant speed.
The object travels with a negative velocity for 10 seconds and then travels at a constant velocity.
The object is slowing down for the first 5 seconds. Then the object is speeding up for the next 5 seconds. Then the object travels at a constant velocity.
The object travels in the negative direction for 10 seconds and then does not move.
The object travels in the positive direction for 5 seconds. Then the object travels in the negative direction at the same speed for the next 5 seconds. Then the object stops moving.
The object is slowing down for the first 5 seconds. Then the object is speeding up for the next 5 seconds. Then the object travels at a constant velocity.
This is a velocity versus time graph. During the first 5 seconds, the velocity is approaching 0. This means that the object is slowing down.
During the next 5 seconds, the velocity is increasing in the negative direction. This means that the object is speeding up in the negative direction.
During the final 5 seconds, the velocity value does not change. This means that the object is remaining at a constant velocity.
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