All AP Physics 1 Resources
Example Questions
Example Question #297 : Ap Physics 1
Using the coordinate system in the graph shown, what is Boomer 's displacement between ?
north
north
None of these
north
north
north
Displacement has magnitude and direction. It is found by subtracting the position at time zero from the position at time ten seconds.
Example Question #21 : Motion In One Dimension
Boomer stands in the center of a long narrow dog run. Boomer walks north in , turns and walks south in , walks another south in , stands still for . What is Boomer's total displacement during these ten seconds?
north
north
None of these
north
north
north
Displacement has direction and magnitude. To determine the displacement, subtract the final position from the initial position, and keep the sign, which shows us direction. Alternatively, we can correctly say that Boomer's displacement is south.
Example Question #22 : Motion In One Dimension
Suppose that an object is dropped from an initial height of 100m above the ground. Neglecting air resistance, how long will it take for this object to reach the ground?
To solve this problem, we'll need to use a formula that can relate distance to acceleration and time. It's also worth noting that in this case, we are only considering vertical motion along the y-axis. We don't have to worry about horizontal motion along the x-axis.
is the vertical displacement
is the initial velocity in the vertical direction
is time
is the acceleration due to gravity (which is only in the vertical direction)
Since the object is starting from rest, the initial velocity will be equal to zero, which cancels out the term and gives us:
Rearranging to solve for , we obtain:
Example Question #21 : Motion In One Dimension
A car traveling at suddenly applies the brakes until it comes to a stop. If the car decelerates at a constant rate of , how long will it take the car to come to a stop?
To answer this question, we'll first need to find the distance the car travels before coming to a stop. Since we are told in the question stem that the car is decelerating at a constant rate, we know that we have a situation in which acceleration is constant and thus we can make use of the kinematic equations.
And since we know that the car is coming to a stop, we know that our final velocity will be equal to zero.
Furthermore, since we know the car is decelerating, we know it is slowing down and is thus accelerating in the direction opposite to its direction of motion. So if we assign a positive value to the velocity, then the acceleration will have a negative value.
Now that we have found the distance traveled by the car, we can calculate the amount of time it takes for the car to come to a stop.
Plug in known values and solve.
From the above expression, we see that there are two solutions that can satisfy the value of . Only one of them, however, will make any sense physically. To solve for the two values, we will need to use the quadratic equation.
Our known values for the variables above are as follows:
Plug into the quadratic equation and solve.
and . The negative value makes no sense physically, thus our answer is .
An alternative (and much faster) way of solving this problem that yields the same answer is to use the following equation.
We know that our final velocity is equal to zero since the car is coming to a stop, and we're told in the question stem that we're dealing with a constant deceleration (negative acceleration).
Example Question #22 : Motion In One Dimension
A hungry wasp spots an fly wandering about. Assuming the wasp attacks the fly from behind (they are both traveling in the same direction) with speed v, and the fly is stationary, what is the speed of the wasp and fly after the collision? Assume the fly and wasp are one object after the collision. Your answer should be in terms of M, m, v where M is the mass of the wasp, m is the mass of the fly and v is the original speed of the wasp.
, they are both stationary after the collision.
Considering the wasp aims to eat the fly, we assume the fly and wasp are one body after the collision. This is an inelastic collision. We can solve this with conservation of momentum.
or
For the two body inelastic colision between the wasp and the fly, we can rewrite this as:
Then taking into account the fact the fly is stationary initially:
Then solve for the velocity of the fly and the wasp after the collision:
Example Question #24 : Motion In One Dimension
A person travelling at a rate of , with initial position at will have travelled to in how much time?
This is a simple question of rate, time, and distance.
, where is distance travelled, is rate, is time passed.
In our case, we know the rate is
We also know that the person travelled to having originally started at at . Keeping this in mind, the distance travelled is:
Now we just solve for time:
Example Question #22 : Motion In One Dimension
Suppose that a ball is thrown straight upward and falls back to the ground in a time . If this same ball is thrown straight upward on a distant planet whose gravity is only one-third that of Earth's, then will change by what factor?
Increase by a factor of
Increase by a factor of
Decrease by a factor of
Decrease by a factor of
Increase by a factor of
In this question, we're being asked to determine how long a ball will remain in the air when it is thrown vertically upward on a planet with reduced gravity. First, we'll need to find an expression that relates gravity with the amount of time the ball remains in the air. To do this, we can make use of the kinematics equations.
Furthermore, since we know the ball will land where it began, we know that .
Moreover, if we define the upward direction as positive and the downward direction as negative, then we know that , since gravity is always pointing in the downward direction.
The above expression is the one we're looking for because it relates time and gravity. From this expression, we can conclude that if the magnitude of gravity is reduced by a factor of three, then the time variable will increase by a factor of three.
Example Question #261 : Newtonian Mechanics
Suppose that a car undergoing uniform acceleration starts from rest and travels a distance of in a time span of . What is the acceleration that this car experiences?
In this problem, we're told that a uniformly accelerating car travels a certain distance in a given amount of time, and we're asked to solve for the acceleration.
For starters, it's important to notice that the car is undergoing uniform acceleration. This means that the car's acceleration is constant, which is important because it means we can utilize the kinematic equations. For this problem, we'll need to use an equation that relates displacement, time, and acceleration.
Because the car is initially starting at rest, we can set the term equal to zero.
Example Question #21 : Motion In One Dimension
A high jumper in track and field jumps from the ground at . How high is she able to jump? Assume gravity is .
To solve this problem we look to the kinematic equations. Based on our need to include distance there are only two equations that might work and since we only know the initial speed we will use:
Because gravity is the only force acting to slow her upward rise we can find the time:
Example Question #25 : Linear Motion And Momentum
Consider a particle initially located at and moving with initial velocity . Assuming a constant acceleration of , calculate the position at a time of .
Looking at the initial information we are given about the particle at , we can construct the equation of motion for the position of the particle as:
Plug in our values and solve.
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