Displacement Velocity And Acceleration

Equation used: $$\vec{d}={x}_{f}-{x}_{i}$$ Given: $${x}_{f}=3\text{ m}$$ $${x}_{i}=8\text{ m}$$ Substitution: $$\vec{d}=3-8$$ $$\vec{d}=-5\text{ m}$$ The displacement is −5 m. The negative sign indicates the object moved 5 m in the negative direction.

The displacement is zero because the runner’s final position is identical to her initial position, so the change in position is zero. The distance travelled is 400 m because distance measures the total path length. Displacement depends only on the start and end positions, while distance accumulates every metre of the path, so they differ whenever the path is not a straight line in one direction.

Converting distance to SI: $$90\text{ km}=90\times 1000=90,000\text{ m}$$ $$30\text{ km}=30\times 1000=30,000\text{ m}$$ Converting time to SI: $$1.5\text{ h}=1.5\times 3600=5400\text{ s}$$ $$0.5\text{ h}=0.5\times 3600=1800\text{ s}$$ Finding total displacement: $$\vec{d}=(+90,000)+(-30,000)=+60,000\text{ m}$$ Finding total time: $$\Delta t=5400+1800=7200\text{ s}$$ Equation used: $${\vec{v}}_{avg}=\frac{\vec{d}}{\Delta t}$$ Substitution: $${\vec{v}}_{avg}=\frac{60,000}{7200}$$ $${\vec{v}}_{avg}=+8.33\text{ m/s}$$ The average velocity is +8.33 m/s (north).

The magnitude of average velocity is less than the average speed. For a round trip the displacement is zero, so average velocity is zero. The total distance is nonzero because the object moved along a path, so average speed is positive. Zero is less than any positive value, so the magnitude of average velocity is less than the average speed.

Equation used: $${\vec{a}}_{avg}=\frac{{\vec{v}}_{f}-{\vec{v}}_{i}}{\Delta t}$$ Given: $${\vec{v}}_{f}=+10.0\text{ m/s}$$ $${\vec{v}}_{i}=+4.0\text{ m/s}$$ $$\Delta t=2.0\text{ s}$$ Substitution: $${\vec{a}}_{avg}=\frac{10.0-4.0}{2.0}$$ $${\vec{a}}_{avg}=\frac{6.0}{2.0}$$ $${\vec{a}}_{avg}=+3.0{\text{ m/s}}^{2}$$ The positive sign confirms the acceleration is in the same direction as the velocity, so the ball speeds up.

The satellite’s acceleration is nonzero. Although its speed is constant, the direction of its velocity continuously changes as it follows the circular path. Because acceleration occurs whenever the magnitude and/or direction of velocity changes, a change in direction alone is sufficient for nonzero acceleration.

The object is speeding up. The velocity and the acceleration are both in the negative direction (same sign). When velocity and acceleration point in the same direction, the speed increases. The magnitude of velocity grows from 6.0 m/s to a larger value over time.

The student draws a tangent line to the curve at $t=3$ s. The tangent line touches the curve at exactly one point. The student then calculates the slope of that tangent line by choosing two widely spaced points on the line and dividing the rise by the run. That slope equals the instantaneous velocity at $t=3$ s.

The answer cannot be determined without knowing whether the curve is concave up or concave down. If the curve is concave up (accelerating), the instantaneous velocity at $t=5$ s is less than the average velocity over the 0–10 s interval because the object moves slower in the first half and faster in the second half. If the curve is concave down (decelerating), the opposite is true. The shape of the graph must be specified before a comparison can be made.