Precalculus Flashcards: Solving Problems With Vectors And Velocity

Study Solving Problems With Vectors And Velocity in Precalculus with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

Precalculus

Solving Problems With Vectors And Velocity

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QUESTION
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With r0=1,2\vec{r}_0=\langle 1,2\rangle and v=3,1\vec{v}=\langle 3,-1\rangle, what is r(4)\vec{r}(4)?

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ANSWER

13,2\langle 13,-2\rangle. 1,2+43,1=1+12,24\langle 1,2\rangle+4\langle 3,-1\rangle=\langle 1+12,2-4\rangle.

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This deck focuses on Solving Problems With Vectors And Velocity, giving you a quick way to review the definitions, rules, and examples that matter most for Precalculus.

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Flashcard 1: With r0=1,2\vec{r}_0=\langle 1,2\rangle and v=3,1\vec{v}=\langle 3,-1\rangle, what is r(4)\vec{r}(4)?

Answer: 13,2\langle 13,-2\rangle. 1,2+43,1=1+12,24\langle 1,2\rangle+4\langle 3,-1\rangle=\langle 1+12,2-4\rangle.

Flashcard 2: What is the formula for the projection of v\vec{v} onto u\vec{u} (vector projection)?

Answer: projuv=vuu2u\text{proj}_{\vec{u}}\vec{v}=\frac{\vec{v}\cdot\vec{u}}{\|\vec{u}\|^2}\,\vec{u}. Projects v\vec{v} onto u\vec{u} using dot product and scaling.

Flashcard 3: What is the unit vector in the direction of v\vec{v} (assuming v0\vec{v}\neq\vec{0})?

Answer: v^=vv\hat{v}=\frac{\vec{v}}{\|\vec{v}\|}. Divide vector by its magnitude to get length 1.

Flashcard 4: Find proj1,03,4\operatorname{proj}_{\langle 1,0\rangle}\langle 3,4\rangle.

Answer: 3,0\langle 3,0\rangle. Projects onto x-axis: keeps x-component, zeros y.

Flashcard 5: State the formula for the vector projection of a\vec{a} onto a nonzero vector b\vec{b}.

Answer: projba=abb2b\operatorname{proj}_{\vec{b}}\vec{a}=\frac{\vec{a}\cdot\vec{b}}{\|\vec{b}\|^2}\vec{b}. Scales b\vec{b} by the scalar projection ratio.

Flashcard 6: What is the formula for the scalar component of v\vec{v} in the direction of u\vec{u}?

Answer: compuv=vuu\text{comp}_{\vec{u}}\vec{v}=\frac{\vec{v}\cdot\vec{u}}{\|\vec{u}\|}. Gives the signed length of the projection of v\vec{v} onto u\vec{u}.

Flashcard 7: Find position at t=3t=3 if r0=2,1\vec{r}_0=\langle 2,-1\rangle and v=4,5\vec{v}=\langle -4,5\rangle.

Answer: 10,14\langle -10,\,14\rangle. r(3)=2,1+34,5=212,1+15\vec{r}(3) = \langle 2,-1\rangle + 3\langle -4,5\rangle = \langle 2-12, -1+15\rangle.

Flashcard 8: What is the distance traveled in time tt at constant velocity vector v\vec{v} (assume t0t\ge 0)?

Answer: distance=vt\text{distance}=\|\vec{v}\|t. Speed times time equals distance traveled.

Flashcard 9: What is the component form of the displacement vector from (x1,y1)(x_1,y_1) to (x2,y2)(x_2,y_2)?

Answer: x2x1, y2y1\langle x_2-x_1,\ y_2-y_1\rangle. Subtract initial from terminal coordinates to get displacement.

Flashcard 10: State the constant-velocity position formula for a particle with r(0)=r0\vec{r}(0)=\vec{r}_0 and velocity v\vec{v}.

Answer: r(t)=r0+tv\vec{r}(t)=\vec{r}_0+t\vec{v}. Linear motion: start at r0\vec{r}_0, move by tvt\vec{v}.

Flashcard 11: State the formula for the scalar projection of a\vec{a} onto a nonzero vector b\vec{b}.

Answer: compba=abb\operatorname{comp}_{\vec{b}}\vec{a}=\frac{\vec{a}\cdot\vec{b}}{\|\vec{b}\|}. Measures signed length of a\vec{a}'s shadow on b\vec{b}.

Flashcard 12: What condition on uv\vec{u}\cdot\vec{v} shows two nonzero vectors are perpendicular?

Answer: uv=0\vec{u}\cdot\vec{v}=0. Perpendicular vectors have a dot product of zero.

Flashcard 13: Identify the condition for two vectors a\vec{a} and b\vec{b} to be perpendicular using a dot product.

Answer: ab=0\vec{a}\cdot\vec{b}=0. Perpendicular vectors have dot product zero.

Flashcard 14: State the formula for scalar multiplication: ka,bk\langle a,b\rangle.

Answer: ka,kb\langle ka,\,kb\rangle. Multiply each component by the scalar kk.

Flashcard 15: Find a unit vector in the direction of v=3,4\vec{v}=\langle 3,4\rangle.

Answer: 35,45\left\langle \frac{3}{5},\,\frac{4}{5}\right\rangle. v=5\|\vec{v}\| = 5, so v^=153,4\hat{v} = \frac{1}{5}\langle 3,4\rangle.

Flashcard 16: Find the speed of v=6,8\vec{v}=\langle 6,8\rangle.

Answer: 1010. v=62+82=36+64=100=10\|\vec{v}\| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10.

Flashcard 17: State the formula for the magnitude of a 22D vector v=a,b\vec{v}=\langle a,b\rangle.

Answer: v=a2+b2\|\vec{v}\|=\sqrt{a^2+b^2}. Apply the Pythagorean theorem to vector components.

Flashcard 18: What is the position vector formula for constant velocity: initial r0\vec{r}_0 and velocity v\vec{v}?

Answer: r(t)=r0+tv\vec{r}(t)=\vec{r}_0+t\vec{v}. Position equals initial position plus displacement over time.

Flashcard 19: What is the magnitude of a 22D vector v=a,b\vec{v}=\langle a,b\rangle?

Answer: v=a2+b2\|\vec{v}\|=\sqrt{a^2+b^2}. Apply the Pythagorean theorem to find the length of the vector.

Flashcard 20: Find the speed (magnitude) of the velocity vector 5,12\langle 5,12\rangle.

Answer: 1313. 52+122=25+144=169=13\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13.

Flashcard 21: Compute the dot product 2,13,4\langle 2,-1\rangle\cdot\langle 3,4\rangle.

Answer: 22. (2)(3)+(1)(4)=64=2(2)(3)+(-1)(4)=6-4=2.

Flashcard 22: What is the formula for the angle θ\theta between nonzero vectors using the dot product?

Answer: cosθ=abab\cos\theta=\frac{\vec{a}\cdot\vec{b}}{\|\vec{a}\|\,\|\vec{b}\|}. Rearranged from ab=abcosθ\vec{a}\cdot\vec{b}=\|\vec{a}\|\|\vec{b}\|\cos\theta.

Flashcard 23: State the dot product formula for a=a1,a2\vec{a}=\langle a_1,a_2\rangle and b=b1,b2\vec{b}=\langle b_1,b_2\rangle.

Answer: ab=a1b1+a2b2\vec{a}\cdot\vec{b}=a_1b_1+a_2b_2. Multiply corresponding components and add.

Flashcard 24: Find the magnitude of v=3,4\vec{v}=\langle 3,4\rangle.

Answer: 55. 32+42=9+16=25=5\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5.

Flashcard 25: State the formula for vector addition in components: a,b+c,d\langle a,b\rangle+\langle c,d\rangle.

Answer: a+c,b+d\langle a+c,\,b+d\rangle. Add corresponding components to get the resultant vector.

Flashcard 26: Identify whether 1,2\langle 1,2\rangle and 4,2\langle 4,-2\rangle are perpendicular.

Answer: Yes, since 1,24,2=0\text{Yes, since }\langle 1,2\rangle\cdot\langle 4,-2\rangle=0. (1)(4)+(2)(2)=44=0(1)(4) + (2)(-2) = 4 - 4 = 0, so they are perpendicular.

Flashcard 27: What is the component form of the vector from A(x1,y1)A(x_1,y_1) to B(x2,y2)B(x_2,y_2)?

Answer: x2x1,y2y1\langle x_2-x_1,\,y_2-y_1\rangle. Subtract initial coordinates from terminal coordinates to get components.

Flashcard 28: What is the unit vector in the direction of a nonzero vector v\vec{v}?

Answer: v^=vv\hat{v}=\frac{\vec{v}}{\|\vec{v}\|}. Divide the vector by its magnitude to get a vector of length 1.

Flashcard 29: Compute the dot product 2,15,4\langle 2,-1\rangle\cdot\langle 5,4\rangle.

Answer: 66. (2)(5)+(1)(4)=104=6(2)(5) + (-1)(4) = 10 - 4 = 6.

Flashcard 30: What is the dot product formula for u=a,b\vec{u}=\langle a,b\rangle and v=c,d\vec{v}=\langle c,d\rangle?

Answer: uv=ac+bd\vec{u}\cdot\vec{v}=ac+bd. Multiply corresponding components and add the products.

Flashcard 31: What is the displacement vector after time tt with constant velocity v\vec{v}?

Answer: Δr=tv\Delta\vec{r}=t\vec{v}. Displacement equals velocity times time for constant motion.

Flashcard 32: Find the resultant of perpendicular velocities 5,0\langle 5,0\rangle and 0,12\langle 0,12\rangle.

Answer: 5,12\langle 5,12\rangle. Add components: 5+0,0+12=5,12\langle 5+0,0+12\rangle=\langle 5,12\rangle.

Flashcard 33: Identify the speed if velocity is v=vx,vy\vec{v}=\langle v_x,v_y\rangle.

Answer: speed=v=vx2+vy2\text{speed}=\|\vec{v}\|=\sqrt{v_x^2+v_y^2}. Speed is the magnitude of the velocity vector.

Flashcard 34: Find projuv\text{proj}_{\vec{u}}\vec{v} for u=1,0\vec{u}=\langle 1,0\rangle and v=3,4\vec{v}=\langle 3,4\rangle.

Answer: 3,0\langle 3,\,0\rangle. vuu2u=311,0=3,0\frac{\vec{v}\cdot\vec{u}}{\|\vec{u}\|^2}\vec{u} = \frac{3}{1}\langle 1,0\rangle = \langle 3,0\rangle.

Flashcard 35: Find the resultant velocity: 3,2+5,7\langle 3, -2\rangle+\langle -5, 7\rangle.

Answer: 2,5\langle -2,\,5\rangle. Add components: (35,2+7)=2,5(3-5, -2+7) = \langle -2, 5\rangle.

Flashcard 36: What is the angle relation for dot product using magnitudes and angle θ\theta?

Answer: uv=uvcosθ\vec{u}\cdot\vec{v}=\|\vec{u}\|\,\|\vec{v}\|\cos\theta. Relates dot product to magnitudes and the angle between vectors.

Flashcard 37: A boat has velocity 4,0\langle 4,0\rangle in still water; current is 0,3\langle 0,3\rangle. Find ground velocity.

Answer: 4,3\langle 4,3\rangle. Add boat and current vectors component-wise.