Distance and Displacement Numerical Problems, MCQs, Q&A

ANAND CLASSES study material and notes explore Numerical Problems and multiple-choice questions (MCQs) on distance and displacement with detailed answers and step-by-step explanations. Learn key concepts with solved examples using mathematical equations

Problem.1

A man travels in the following sequence:
➡ 1.5 m East
⬇ 2.0 m South
➡ 4.5 m East

(i) What is the total distance traveled?

(ii) What is the resultant displacement?

🚀 CBSE 🚀

Solution :

Total distance = 1.5 + 2.0 + 4.5 = 8.0 m

To determine the resultant displacement, we trace his path on a diagram.

• Draw 1.5 m East (AB)
• Draw 2.0 m South (BC)
• Draw 4.5 m East (CD)

Now, the resultant displacement is represented by a straight line from the starting point A to the final point D. Measuring the length of AD, we get 6.3 m.

AD² = (AB + CD)² + BC²

AD² = (1.5 + 4.5)² + 2²

AD = 6.3 m.

🔹 Final displacement = 6.3 m
🔹 Key Concept: Displacement is the shortest straight-line distance between the starting and ending points, while distance is the total path traveled.

🚀 Did You Know?

• Distance is a scalar quantity (magnitude only), whereas displacement is a vector quantity (magnitude + direction).
• Even if a person moves a lot, their displacement could be zero if they return to the starting point!

Problem.2

A man walks 10 meters towards the north and then 20 meters towards the east. What is his displacement?

(a) 22.5 m
(b) 25 m
(c) 25.5 m
(d) 30 m

🚀 AIPMT, JIPMER & AFMC Question 🚀

Solution :

Concept of Displacement : Displacement is the shortest straight-line distance between the initial and final positions of an object. It differs from distance, which is the total path traveled. Displacement is a vector quantity, meaning it has both magnitude and direction.

Given, a man moves 10m North and then 20m East.

👉 Taking East as x-axis and North as y-axis, the displacement vector d is:

\begin{array}{l} \mathbf{d} = 20 \hat{i} + 10 \hat{j} \end{array}

📐 Magnitude of displacement: \begin{aligned}
\text{|d|} &=\begin{array}{l} \sqrt{20^2 + 10^2} = \sqrt{400 + 100} = \sqrt{500} = 10\sqrt{5} \approx 22.5 \text{m} \end{array}\end{aligned}

✅ Correct answer: (a) 22.5m

Problem.3

🚀 Distance & Displacement in Circular Motion 🚀

A body moves over one-fourth of a circular arc in a circle of radius r. What are the distance traveled and displacement?

Solution:

Let particle start from A, its position vector r_A and after one quarter position vector r_B

✅ Position vectors:

• Initial position at A:

\begin{array}{l} \mathbf{r}_A = r \hat{i} \end{array}

• Final position at B (after 90° movement):

\begin{array}{l} \mathbf{r}_B = r \hat{j} \end{array}

✅ Displacement: \begin{array}{l} \mathbf{d} = \mathbf{r}_B – \mathbf{r}_A = r\hat{j} – r\hat{i} \end{array}

📐 Magnitude of displacement: \begin{array}{l} |\mathbf{d}| = \sqrt{r^2 + r^2} = r\sqrt{2} \end{array}

✅ Distance traveled (arc length): \begin{array}{l} \text{Distance} = \frac{1}{4} \times 2\pi r = \frac{\pi r}{2} \end{array}

Final Answer:

✅ Correct option: (a) πr/2, r√2

🚀 Displacement of a Point on a Rolling Wheel 🚀

Problem.4

When a wheel rolls forward by half a revolution, what is the displacement of the initial contact point with the ground? The radius of the wheel is R.

🚀 AIEEE, NDA 🚀

Solution:

✅ Horizontal distance covered by the wheel in half a revolution: \begin{array}{l} \text{Distance} = \pi R \end{array}

✅ Vertical displacement of the point (moves from bottom to top of the wheel): \begin{array}{l} \text{Vertical shift} = 2R \end{array}

✅ Total displacement (using Pythagoras theorem): \begin{array}{l} d = \sqrt{(\pi R)^2 + (2R)^2} \end{array} \begin{array}{l} d = R \sqrt{\\π^2 + 4} \end{array}

Final Answer:

✅ Correct option: (b) R√(π² + 4)

Problem 1: Motion in a Straight Line

A car moves 600 m east and then 400 m west. Find:

1. The total distance traveled
2. The displacement

Solution:

Total distance traveled: \begin{array}{l} S = 600 + 400 = 1000 \text{ m} \end{array}

Displacement: \begin{array}{l} D = 600 – 400 = 200 \text{ m} \quad (\text{towards east}) \end{array}

Problem 2: Circular Motion

A runner moves along a circular track of radius 14 m and completes half a revolution. Find:

1. The distance traveled
2. The displacement

Solution:

Distance traveled: \begin{array}{l} S = \pi r = \pi (14) = 44 \text{ m} \quad (\text{using} \, \pi \approx 3.14) \end{array}

Displacement (diameter of the circle): \begin{array}{l} D = 2r = 2(14) = 28 \text{ m} \end{array}

Problem 3: Motion in Two Dimensions

A person walks 6 m east and then 8 m north. Find:

1. The total distance traveled
2. The displacement

Solution:

Total distance traveled: \begin{array}{l} S = 6 + 8 = 14 \text{ m} \end{array}

Displacement using the Pythagorean theorem: \begin{array}{l} D = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ m} \end{array}

Distance and Displacement – MCQs with Answers & Explanations

1. A car moves 40 m east and then 30 m west. What is the distance and displacement?

(a) 70 m, 10 m east
(b) 10 m, 70 m west
(c) 70 m, 10 m west
(d) 10 m, 70 m east

✅ Answer: (a) 70 m, 10 m east

📝 Explanation: \begin{array}{l} S = 40 + 30 = 70 \text{ m} \end{array} \begin{array}{l} D = 40 – 30 = 10 \text{ m (east)} \end{array}

2. If an object moves in a circular path and returns to the starting point, what is the displacement?

(a) Equal to the circumference of the circle
(b) Equal to the diameter of the circle
(c) Zero
(d) None of the above

✅ Answer: (c) Zero

📝 Explanation: \begin{array}{l} \text{Distance} = 2 \pi r \end{array} \begin{array}{l} \text{Displacement} = 0 \text{ (since initial and final positions are the same)} \end{array}

3. A person walks 6 m north and then 8 m east. What is the displacement?

(a) 10 m
(b) 14 m
(c) 2 m
(d) 12 m

✅ Answer: (a) 10 m

📝 Explanation: \begin{array}{l} S = 6 + 8 = 14 \text{ m} \end{array} \begin{array}{l} D = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ m} \end{array}

4. A man walks 3 m north, 4 m east, and then 3 m south. What is the displacement?

(a) 10 m, 4 m east
(b) 7 m, 3 m west
(c) 7 m, 3 m east
(d) 10 m, 3 m west

✅ Answer: (a) 10 m, 4 m east

📝 Explanation: \begin{array}{l} S = 3 + 4 + 3 = 10 \text{ m} \end{array} \begin{array}{l} D = 4 \text{ m (east, as north-south cancels out)} \end{array}

5. Which statement is always true about displacement?

(a) It is always greater than or equal to distance
(b) It is always positive
(c) It is always less than or equal to distance
(d) It is always equal to distance

✅ Answer: (c) It is always less than or equal to distance

📝 Explanation: \begin{array}{l} \text{Displacement} \leq \text{Distance} \end{array}

6. A car moves 5 km north, then 5 km east, and finally 5 km south. What is its displacement?

(a) 5 km
(b) 10 km
(c) 0 km
(d) 525\sqrt{2} km

✅ Answer: (d) 525\sqrt{2} km

📝 Explanation: \begin{array}{l} S = 5 + 5 + 5 = 15 \text{ km} \end{array} \begin{array}{l} D = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \text{ km} \end{array}

7. If the displacement of a moving object is zero, what can we conclude?

(a) The object has not moved
(b) The distance traveled is also zero
(c) The object has returned to its starting point
(d) The object moved in a straight line

✅ Answer: (c) The object has returned to its starting point

📝 Explanation: \begin{array}{l} D = 0 \Rightarrow \text{Initial Position} = \text{Final Position} \end{array}

8. A particle moves 100 m forward and then 40 m backward. What is the displacement?

(a) 140 m
(b) 60 m
(c) -60 m
(d) 100 m

✅ Answer: (b) 60 m

📝 Explanation: \begin{array}{l} S = 100 + 40 = 140 \text{ m} \end{array} \begin{array}{l} D = 100 – 40 = 60 \text{ m (forward)} \end{array}

9. When an object moves in a straight line without changing direction, what is true about distance and displacement?

(a) Distance is greater than displacement
(b) Distance is less than displacement
(c) Distance is equal to displacement
(d) Cannot be determined

✅ Answer: (c) Distance is equal to displacement

📝 Explanation: \begin{array}{l} \text{If no change in direction, then } S = D \end{array}

10. A ball is thrown vertically upwards with speed 20 m/s, reaches the highest point, and returns to the starting position. What is the displacement?

(a) 40 m
(b) 20 m
(c) 0 m
(d) 10 m

✅ Answer: (c) 0 m

📝 Explanation: \begin{array}{l} \text{Since the ball returns to the same position, } D = 0 \end{array}

🏫 Best Coaching Center for IIT-JEE, NEET, and Foundations – ANAND CLASSES

📚 Buy complete study material: 📖 Anand Classes Study Material
🏫 Proprietor: NIRMAL ANAND Educations
✍️ Written by: Neeraj Anand
📖 Published by: Anand Technical Publishers Under Anand Classes
📞 Contact: 📲 +91-9463138669
📧 Email: 📩 anandclasses1996@gmail.com