Scalars And Vectors

Find the magnitude: $$|\vec{F}|=\sqrt{(-8{)}^{2}+{6}^{2}}$$ $$|\vec{F}|=\sqrt{64+36}=\sqrt{100}=10\text{ N}$$ Find the reference angle: $${\theta }_{\text{ref}}={tan}^{-1}\left(\frac{6}{8}\right)=36.9^{\circ}$$ The x-component is negative and the y-component is positive. This places the vector in the second quadrant. $$\theta =180^{\circ}-36.9^{\circ}=143^{\circ}\text{ from the +x axis}$$ The force has a magnitude of 10 N directed at 143° from the positive x-axis.

Distance is greater. Distance is a scalar that accumulates every meter traveled, so the total distance is 10 km. Displacement is a vector from start to finish, and because the car returns to its starting point, the displacement is 0 km. A scalar sum of path lengths is always greater than or equal to the magnitude of the net displacement.

Starting point: The resultant vector equals the component-wise sum of the addend vectors. $$\vec{R}=\vec{A}+\vec{B}$$ $$\vec{R}=(3+(-1))\hat{i}+((-2)+5)\hat{j}$$ $$\vec{R}=2\hat{i}+3\hat{j}\text{ m}$$ Find the magnitude: $$|\vec{R}|=\sqrt{{2}^{2}+{3}^{2}}=\sqrt{4+9}=\sqrt{13}$$ $$|\vec{R}|\approx 3.61\text{ m}$$

The claim is not consistent. Speed and velocity share the same SI unit, but they are fundamentally different. Speed is a scalar - it has magnitude only and is always non-negative. Velocity is a vector - it has both magnitude and direction. A car moving at 20 m/s west and another at 20 m/s east have the same speed but different velocities. The unit alone does not determine whether a quantity is scalar or vector.

At the highest point, velocity is zero — the ball has momentarily stopped before reversing direction. Acceleration is negative. Gravity acts downward, which is the negative direction in this coordinate system. The acceleration due to gravity is $a=-g=-10$ m/s² regardless of the ball’s position or velocity.

Distance is a scalar, so it accumulates without regard to direction: total distance is 11 m, a positive number. Displacement is a vector from start to finish. The object ends 5 m to the right of where it started, so the displacement is +5 m. The positive sign indicates the net displacement is in the positive (rightward) direction. Distance has no sign; displacement carries a sign that shows direction.