Question 1 of 5
Michelle walks 100m toward the west, then turns and walks back the way she came 20m. What is her distance? What is her displacement?
Why: **Solution:**
**Distance:** Distance is the total path length traveled, regardless of direction. Michelle walks 100 m west + 20 m east (back), so total distance = 100 m + 20 m = 120 m.
**Displacement:** Displacement is the straight-line distance from initial to final position with direction. Starting at origin, after 100 m west: position = -100 m. Then 20 m back (east): final position = -100 m + 20 m = -80 m. Thus, displacement = 80 m west.
**Key Difference:** Distance is scalar (120 m), displacement is vector (80 m west). [2]
Question 2 of 5
A football field is about 100 m long. If it takes a person 20 seconds to run its length, how fast (what speed) were they running?
Why: **Solution:**
Speed is defined as total distance traveled divided by total time taken. Here, distance = 100 m, time = 20 s.
Speed = \( \frac{\text{distance}}{\text{time}} = \frac{100\,\text{m}}{20\,\text{s}} = 5\,\text{m/s} \).
This is average speed since it's total distance over total time. Note: If it were displacement (straight line, same as distance here), velocity would also be 5 m/s east (assuming direction from start to end). [8]
Question 3 of 5
A mountain climbing expedition establishes a base camp and two intermediate camps, A and B. Camp A is 11,200 m east of and 3,200 m above base camp. Camp B is 8400 m east of and 1700 m higher than Camp A. Determine the displacement between base camp and Camp B.
Why: **Solution:**
Displacement is the straight-line vector from base camp to Camp B.
**Position of Camp A relative to base:** \( \vec{d_A} = 11,200\,\hat{i} + 3,200\,\hat{j} \) m
**Position of Camp B relative to Camp A:** \( \vec{d_B-A} = 8,400\,\hat{i} + 1,700\,\hat{j} \) m
**Total displacement:** \( \vec{d_{B}} = \vec{d_A} + \vec{d_B-A} = (11,200 + 8,400)\,\hat{i} + (3,200 + 1,700)\,\hat{j} = 19,600\,\hat{i} + 4,900\,\hat{j} \) m
**Magnitude:** \( |\vec{d_B}| = \sqrt{(19,600)^2 + (4,900)^2} = \sqrt{384,160,000 + 24,010,000} = \sqrt{408,170,000} \approx 15,100\) m
**Direction:** \( \theta = \tan^{-1}\left(\frac{4,900}{19,600}\right) = \tan^{-1}(0.25) \approx 24.1^\circ \) above east.
Note: Distance traveled via path A-B would be much longer (~22,000 m), but displacement is shortest straight-line path. [9]
Question 4 of 5
Which of the following statements about velocity are correct? (MCQ)
A. Velocity is the distance traveled by an object over an interval of time.
B. Velocity is a scalar quantity with both magnitude and direction.
C. It is possible for an object to travel at constant speed without constant velocity.
D. The magnitude of an object's average velocity is its average speed.
Why: **Analysis of options:**
**A. Incorrect:** Velocity uses displacement (vector), not distance (scalar).
**B. Incorrect:** Velocity is vector (magnitude + direction); scalars have only magnitude.
**C. Correct:** Uniform circular motion: constant speed, changing velocity (direction changes). Example: Car around track at constant 50 km/h - velocity changes continuously.
**D. Incorrect:** Generally not equal. If path is straight, |avg velocity| = avg speed. But circular path: avg velocity = 0, avg speed > 0.
**Key Concepts:** Speed = distance/time (scalar), Velocity = displacement/time (vector). [7]
Question 5 of 5
A car drives east at 40 m/s for 500 seconds, stops, turns around, then drives west at 50 m/s for 400 seconds and stops. What is the displacement of the car?
Why: **Solution:**
Displacement is net change in position (vector).
**Eastward leg:** Distance = 40 m/s × 500 s = 20,000 m east (displacement = +20,000 m)
**Westward leg:** Distance = 50 m/s × 400 s = 20,000 m west (displacement = -20,000 m)
**Total displacement:** +20,000 m + (-20,000 m) = 0 m
**Total distance traveled:** 20,000 m + 20,000 m = 40,000 m
**Average velocity:** \( v_{avg} = \frac{\Delta x}{\Delta t} = \frac{0}{900} = 0\,\text{m/s} \)
**Average speed:** \( \frac{40,000}{900} \approx 44.4\,\text{m/s} \)
Car returns to start, so displacement = 0 despite distance = 40 km. [3]
