AP Physics 1 : Newtonian Mechanics

Study concepts, example questions & explanations for AP Physics 1

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Example Questions

Example Question #251 : Newtonian Mechanics

Boomerstablevt

According to the table above, during which interval does Boomer reach his highest speed?

Possible Answers:

\(\displaystyle 2.5s\rightarrow5s\)

\(\displaystyle 0s\rightarrow2.5s\)

\(\displaystyle 7.5s\rightarrow10s\)

\(\displaystyle 5s\rightarrow7.5s\)

 None: there is a tie

Correct answer:

 None: there is a tie

Explanation:

Boomer's speed is the ratio of the distance traveled to the time. It is \(\displaystyle 2\frac{m}{s}\) during both the first and and third intervals.

Example Question #291 : Ap Physics 1

Boomervt1

According to the graph above, Boomer has the slowest speed during which interval?

Possible Answers:

\(\displaystyle 0s\rightarrow 2.5s\)

\(\displaystyle 5s \rightarrow 7.5s\)

\(\displaystyle 2.5s\rightarrow 5s\)

\(\displaystyle 7.5s\rightarrow 10s\)

None of these

Correct answer:

\(\displaystyle 7.5s\rightarrow 10s\)

Explanation:

The smallest speed is zero which occurs during the final intervals, which have a slope of zero.

Example Question #292 : Ap Physics 1


Boomerstablevt

According to the graph above, when does Boomer have the smallest average speed?

Possible Answers:

\(\displaystyle 7.5s\rightarrow 10s\)

\(\displaystyle 2.5s\rightarrow5s\)

\(\displaystyle 5s\rightarrow 7.5s\)

None: there is a tie

\(\displaystyle 0s\rightarrow 2.5s\)

Correct answer:

\(\displaystyle 7.5s\rightarrow 10s\)

Explanation:

The smallest speed is zero, which occurs during the final intervals which shows no change in position.

Example Question #293 : Ap Physics 1

Boomervt1

According to the graph above, Boomer has the largest positive velocity during which interval?

Possible Answers:

\(\displaystyle 0s\rightarrow2.5s\)

\(\displaystyle 7.5s\rightarrow10s\)

None of these

\(\displaystyle 2.5s\rightarrow5s\)

\(\displaystyle 5s\rightarrow7.5s\)

Correct answer:

\(\displaystyle 0s\rightarrow2.5s\)

Explanation:

The largest and only positive velocity occurs during the first interval when the slope is positive.

Example Question #294 : Ap Physics 1

Boomerstablevt

According to the table above, when does Boomer have the largest positive velocity?

 

Possible Answers:

None of these

\(\displaystyle 2.5s\rightarrow5s\)

\(\displaystyle 5s\rightarrow7.5s\)

\(\displaystyle 0s\rightarrow 2.5s\)

\(\displaystyle 7.5\rightarrow10s\)

Correct answer:

\(\displaystyle 0s\rightarrow 2.5s\)

Explanation:

The largest positive velocity occurs during the period with the maximum positive change in position. This is the first period.

Example Question #295 : Ap Physics 1

Boomervt1

According to the graph above, Boomer has the largest negative velocity during which interval?

Possible Answers:

\(\displaystyle 5s\rightarrow7.5s\)

\(\displaystyle 2.5s\rightarrow5s\)

\(\displaystyle 7.5s\rightarrow10s\)

 None of these

\(\displaystyle 0s\rightarrow2.5s\)

Correct answer:

\(\displaystyle 5s\rightarrow7.5s\)

Explanation:

Negative velocities correspond to periods when the position decreases in value. The largest negative velocity occurs during the third period.

Example Question #295 : Ap Physics 1

Boomerstablevt

According to the table above, when does Boomer have his largest negative velocity?

Possible Answers:

\(\displaystyle 0s\rightarrow2.5s\)

\(\displaystyle 2.5s\rightarrow5s\)

None of these

\(\displaystyle 5s\rightarrow7.5s\)

\(\displaystyle 7.5s\rightarrow10s\)

Correct answer:

\(\displaystyle 5s\rightarrow7.5s\)

Explanation:

The position decline the largest amount per unit time during the third interval.

Example Question #296 : Ap Physics 1

Boomervt1

Using the coordinate system in the graph shown, what is Boomer 's displacement between \(\displaystyle 0s\rightarrow10s\)?

Possible Answers:

\(\displaystyle 12.5m\) north

\(\displaystyle 1m\) north

\(\displaystyle 2.5m\) north

\(\displaystyle -2.5m\) north

None of these

Correct answer:

\(\displaystyle -2.5m\) north

Explanation:

Displacement has magnitude and direction. It is found by subtracting the position at time zero from the position at time ten seconds.

Example Question #297 : Ap Physics 1

Boomer stands in the center of a long narrow dog run. Boomer walks \(\displaystyle 5m\) north in \(\displaystyle 2.5s\), turns and walks \(\displaystyle 2.5m\) south in \(\displaystyle 2.5s\), walks another \(\displaystyle 5m\) south in \(\displaystyle 2.5s\), stands still for \(\displaystyle 2.5s\). What is Boomer's total displacement during these ten seconds?

Possible Answers:

\(\displaystyle 2.5m\) north

None of these

\(\displaystyle 12.5m\) north

\(\displaystyle -2.5m\) north

\(\displaystyle 1m\) north

Correct answer:

\(\displaystyle -2.5m\) north

Explanation:

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 \(\displaystyle 2.5m\) south. 

Example Question #21 : 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?

\(\displaystyle g=9.8\frac{m}{s^2}\)

Possible Answers:

\(\displaystyle 10.20s\)

\(\displaystyle 3.19s\)

\(\displaystyle 4.52s\)

\(\displaystyle 20.41s\)

Correct answer:

\(\displaystyle 4.52s\)

Explanation:

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.

\(\displaystyle \Delta y=v_{i}t+\frac{1}{2}at^{2}\)

\(\displaystyle \Delta y\) is the vertical displacement

\(\displaystyle v_{i}\) is the initial velocity in the vertical direction

\(\displaystyle t\) is time

\(\displaystyle a\) 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 \(\displaystyle v_{i}t\) term and gives us:

\(\displaystyle \Delta y=\frac{1}{2}at^{2}\)

Rearranging to solve for \(\displaystyle t\), we obtain:

\(\displaystyle t=\sqrt{\frac{2\Delta y}{a}}\)

\(\displaystyle t=\sqrt{\frac{2(100 m)}{9.8 \frac{m}{s^{2}}}}=4.52s\)

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