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AP Physics 1 : AP Physics 1

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Example Questions

Example Question #21 : Motion In Two Dimensions

A ball is thrown at a velocity $\vec{v}=\left (10\hat{i}+6\hat{j} \right )\frac{m}{s}$ . What is the speed of the ball?

Possible Answers:

$10\frac{m}{s}$

$4\frac{m}{s}$

$11.66\frac{m}{s}$

$6\frac{m}{s}$

Correct answer:

$11.66\frac{m}{s}$

Explanation:

If a vector is in the form $\vec{r}=a\hat{i}+b\hat{j}$ , the magnitude of a vector, $\left | \vec{r} \right |$ ,is found by using the equation $\left |\vec{r} \right |=\sqrt{a^{2}+b^{2}}$ .

Speed is the magnitude of the velocity vector. Given the velocity vector $\vec{v}=\left (10\hat{i}+6\hat{j} \right )\frac{m}{s}$ , the speed of the ball is:

$\left |\vec{v} \right |=\sqrt{10^{2}+6^{2}}=\sqrt{136}\approx11.66\frac{m}{s}$

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Example Question #21 : Motion In Two Dimensions

An ant is digging at a velocity $\vec{v}=\left (7\hat{i}-5\hat{j} \right )\frac{mm}{s}$ into the Earth. What is the speed of the ant?

Possible Answers:

$5\frac{mm}{s}$

$8.60\frac{mm}{s}$

$7\frac{mm}{s}$

$2\frac{mm}{s}$

Correct answer:

$8.60\frac{mm}{s}$

Explanation:

If a vector is in the form $\vec{r}=a\hat{i}+b\hat{j}$ , the magnitude of a vector, $\left | \vec{r} \right |$ ,is found by using the equation $\left |\vec{r} \right |=\sqrt{a^{2}+b^{2}}$ .

Speed is the magnitude of the velocity vector. Given the velocity vector:

$\vec{v}=\left (7\hat{i}-5\hat{j} \right )\frac{mm}{s}$

The speed of the ant is:

$\left |\vec{v} \right |=\sqrt{7^{2}+(-5)^{2}}=\sqrt{49+25}=\sqrt{74}\approx8.60\frac{mm}{s}$

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Example Question #111 : Linear Motion And Momentum

An ant is digging at a velocity $\vec{v}=\left (7\hat{i}-5\hat{j} \right )\frac{mm}{s}$ in the Earth. In what direction is the ant digging?

Possible Answers:

$59^{\circ}$ right of vertical

$59^{\circ}$ left of vertical

$54.5^{\circ}$ below the horizon

$54.5^{\circ}$ above the horizon

Correct answer:

$54.5^{\circ}$ below the horizon

Explanation:

A two-dimensional vector comes in the form $\vec{r}=a\hat{i}+b\hat{j}$ . The direction of the vector is found using the equation:

$\theta=tan^{-1}\left ( \frac{b}{a}\right )$

For this problem:

$\theta=tan^{-1}\left ( \frac{-7}{5}\right )\approx-54.5^{\circ}$

The ant is digging at an angle $54.5^{\circ}$ below the horizon.

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Example Question #21 : Motion In Two Dimensions

A rollercoaster car is sliding down an incline at a velocity $\vec{v}=\left (-15\hat{i}-25\hat{j} \right )\frac{m}{s}$ . What is the speed of the rollercoaster car?

Possible Answers:

$10\frac{m}{s}$

$25\frac{m}{s}$

$29.15\frac{m}{s}$

$15\frac{m}{s}$

Correct answer:

$29.15\frac{m}{s}$

Explanation:

If a vector is in the form $\vec{r}=a\hat{i}+b\hat{j}$ , the magnitude of a vector, $\left | \vec{r} \right |$ ,is found by using the equation $\left |\vec{r} \right |=\sqrt{a^{2}+b^{2}}$ .

Speed is the magnitude of the velocity vector. Given the velocity vector:

$\vec{v}=\left (-15\hat{i}-25\hat{j} \right )\frac{m}{s}$

The speed of the rollercoaster car is:

$\left |\vec{v} \right |=\sqrt{(-15)^{2}+(-25)^{2}}=\sqrt{225+625}=\sqrt{850}\approx29.15\frac{m}{s}$

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Example Question #21 : Motion In Two Dimensions

A car travels east, south, then west. What is the magnitude of the displacement of the car?

Possible Answers:

8.54m

11m

Correct answer:

8.54m

Explanation:

To find the magnitude of the displacement of the car, we must represent each velocity of the car as a vector. We will then add the vectors. Finally, we will find the magnitude of the resulting velocity vector, which will give us speed.

The car first travels east , represented as a vector: $\vec{r}=\left (7\hat{i} \right )m$

The car then travels south , represented as a vector: $\vec{r}=\left (-8\hat{j} \right )m$

The car finally travels west , represented as a vector: $\vec{r}=\left (-4\hat{i} \right )m$

Adding the vectors together gives us the displacement of the car:

$\vec{r}=\left (7\hat{i}-8\hat{j}-4\hat{i} \right )m=\left (3\hat{i}-8\hat{j} \right )m$

If a vector is in the form $\vec{r}=a\hat{i}+b\hat{j}$ , its magnitude, $\left | \vec{r} \right |$ , is found by using the equation $\left |\vec{r} \right |=\sqrt{a^{2}+b^{2}}$

Given the displacement vector: $\vec{r}=\left (3\hat{i}-8\hat{j} \right )m$ , the magnitude of the displacement vector is:

$\left |\vec{r} \right |=\sqrt{3^{2}+(-8)^{2}}=\sqrt{9+64}=\sqrt{73}\approx8.54m$

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Example Question #22 : Motion In Two Dimensions

A car travels east, south, then west. What is the direction of the displacement of the car?

Possible Answers:

$20.6^{\circ}$ below horizontal

$20.6^{\circ}$ above horizontal

$20.6^{\circ}$ left of vertical

$20.6^{\circ}$ right of vertical

Correct answer:

$20.6^{\circ}$ below horizontal

Explanation:

A two-dimensional vector comes in the form $\vec{r}=a\hat{i}+b\hat{j}$ . The angle of the vector is found using the equation:

$\theta=tan^{-1}\left ( \frac{b}{a}\right )$

For this problem, the displacement vector is:

$\vec{r}=\left (3\hat{i}-8\hat{j} \right )m$

The angle of displacement is:

$\theta=tan^{-1}\left ( \frac{3}{-8}\right )\approx-20.6^{\circ}$

The car is traveling $20.6^{\circ}$ below horizontal.

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Example Question #22 : Motion In Two Dimensions

A boat is crossing a river is traveling east at $7\frac{m}{s}$ . The current of the river is traveling north at $3\frac{m}{s}$ . At what angle is the boat traveling?

Possible Answers:

$82.4^{\circ}$ below horizontal

$23.2^{\circ}$ above horizontal

$82.4^{\circ}$ above horizontal

$23.2^{\circ}$ below horizontal

Correct answer:

$23.2^{\circ}$ above horizontal

Explanation:

A two-dimensional vector comes in the form $\vec{r}=a\hat{i}+b\hat{j}$ . The angle of the vector is found using the equation:

$\theta=tan^{-1}\left ( \frac{b}{a}\right )$

For this problem, the velocity vector is: $\vec{v}=\left (7\hat{i}+3\hat{j} \right )\frac{m}{s}$

The angle of the boat is:

$\theta=tan^{-1}\left ( \frac{3}{7}\right )\approx23.2^{\circ}$

The boat is traveling $23.2^{\circ}$ above horizontal.

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Example Question #21 : Motion In Two Dimensions

A ball is thrown at a velocity $\vec{v}=\left (10\hat{i}+6\hat{j} \right)\frac{m}{s}$ . What is the direction of the ball (the angle at which the ball is thrown)?

Possible Answers:

$59^{\circ}$ above the horizon

$9^{\circ}$ left of vertical

$59^{\circ}$ below the horizon

$9^{\circ}$ right of vertical

Correct answer:

$59^{\circ}$ above the horizon

Explanation:

A two-dimensional vector comes in the form $\vec{r}=a\hat{i}+b\hat{j}$ . The direction of the vector is found using the equation:

$\theta=tan^{-1}\left ( \frac{b}{a}\right )$

For this problem:

$\theta=tan^{-1}\left ( \frac{10}{6}\right )\approx59^{\circ}$

The ball is thrown $59^{\circ}$ above the horizon.

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Example Question #121 : Linear Motion And Momentum

A rollercoaster car is traveling along an incline at a velocity $\vec{v}=\left (-15\hat{i}-25\hat{j} \right )\frac{m}{s}$ . What is the direction of the rollercoaster car?

Possible Answers:

To the right at an angle $54.5^{\circ}$ above the horizon

To the right at an angle $54.5^{\circ}$ below the horizon

To the left at an angle $54.5^{\circ}$ above the horizon

To the left at an angle $54.5^{\circ}$ below the horizon

Correct answer:

To the right at an angle $54.5^{\circ}$ below the horizon

Explanation:

A two-dimensional vector comes in the form $\vec{r}=a\hat{i}+b\hat{j}$ . The direction of the vector is found using the equation:

$\theta=tan^{-1}\left ( \frac{b}{a}\right )$

For this problem:

$\theta=tan^{-1}\left ( \frac{-25}{-15}\right )\approx59^{\circ}$

The rollercoaster car is traveling backwards at an angle $54.5^{\circ}$ below the horizon.

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Example Question #31 : Motion In Two Dimensions

Two cyclist leave from the same starting point. One cyclist is traveling east at $5\frac{m}{s}$ . The other cyclist is traveling south at $4\frac{m}{s}$ . At what speed are the cyclists traveling away from each other?

Possible Answers:

$6.40\frac{m}{s}$

$5\frac{m}{s}$

$9\frac{m}{s}$

$4.2\frac{m}{s}$

Correct answer:

$6.40\frac{m}{s}$

Explanation:

To find the speed at which the cyclists are traveling away from each other, we must represent the velocity of each cyclists as a vector. We will then add the two vectors. Finally, we will find the magnitude of the resulting velocity vector, which will give us speed.

One cyclist is traveling east at $5\frac{m}{s}$ , represented as a vector: $\vec{v}=\left (5\hat{i} \right )\frac{m}{s}$

The other cyclist is traveling south at $4\frac{m}{s}$ , represented as a vector: $\vec{v}=\left (-4\hat{j} \right )\frac{m}{s}$

Adding the two vectors together, gives us the velocity at which the cyclists are traveling away from each other:

$\vec{v}=\left (5\hat{i}-4\hat{j} \right )\frac{m}{s}$

The speed at which the cyclists are traveling away from each other is the magnitude of the velocity.

If a vector is in the form $\vec{r}=a\hat{i}+b\hat{j}$ , its magnitude, $\left | \vec{r} \right |$ ,is found by using the equation $\left |\vec{r} \right |=\sqrt{a^{2}+b^{2}}$

Given the velocity vector:

$\vec{v}=\left (5\hat{i}-4\hat{j} \right )\frac{m}{s}$

The speed at which the cyclists are traveling away from each other is:

$\left |\vec{v} \right |=\sqrt{5^{2}+(-4)^{2}}=\sqrt{25+16}=\sqrt{41}\approx6.40\frac{m}{s}$

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AP Physics 1 : AP Physics 1

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #21 : Motion In Two Dimensions

Example Question #21 : Motion In Two Dimensions

Example Question #111 : Linear Motion And Momentum

Example Question #21 : Motion In Two Dimensions

Example Question #21 : Motion In Two Dimensions

Example Question #22 : Motion In Two Dimensions

Example Question #22 : Motion In Two Dimensions

Example Question #21 : Motion In Two Dimensions

Example Question #121 : Linear Motion And Momentum

Example Question #31 : Motion In Two Dimensions

All AP Physics 1 Resources

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