Distance from Velocity Time Graph Calculation

❓ Question

The velocity-time graph of a body moving in a straight line is shown in the figure.
The point S is at 4.333 seconds.

Find the total distance covered by the body in 6 s.

---

🖼 Question Image

---

✍️ Short Explanation

This problem is based on:

👉 Area under velocity-time graph
👉 Positive and negative velocity
👉 Distance vs displacement concept.

Key idea:

$$\text{Distance}=\text{Sum of absolute areas under }v-t\text{ graph}$$

---

🔷 Step 1 — Area from 0 to 2 s 💯

From graph:

Velocity increases from:

$$0 \rightarrow 4 \text{ m/s}$$

Triangle area:

$$A_1=\frac12 \times 2 \times 4$$
$$A_1=4 \text{ m}$$

---

🔷 Step 2 — Area from 2 to 3 s

Velocity constant:

$$v=4 \text{ m/s}$$

Rectangle area:

$$A_2=1\times4$$
$$A_2=4 \text{ m}$$

---

🔷 Step 3 — Area from 3 to 5 s

Velocity decreases linearly from:

$$4 \rightarrow -2$$

Line crosses axis at point S:

$$t=4.333\text{ s}=\frac{13}{3}\text{ s}$$

Positive area (3 to S)

Base:

$$\frac{13}{3}-3=\frac43$$

Area:

$$A_3=\frac12 \times \frac43 \times 4$$
$$A_3=\frac83 \text{ m}$$

Negative area (S to 5)

Base:

$$5-\frac{13}{3}=\frac23$$

Magnitude of area:

$$A_4=\frac12 \times \frac23 \times 2$$
$$A_4=\frac23 \text{ m}$$

Distance uses absolute value.

---

🔷 Step 4 — Area from 5 to 6 s

Velocity changes from:

$$-2 \rightarrow 0$$

Triangle below axis:

$$A_5=\frac12 \times1\times2$$
$$A_5=1 \text{ m}$$

Distance contribution:

$$1 \text{ m}$$

---

🔷 Step 5 — Total Distance

$$\text{Distance}=A_1+A_2+A_3+A_4+A_5$$
$$=4+4+\frac83+\frac23+1$$
$$=9+\frac{10}{3}$$
$$=\frac{37}{3}\text{ m}$$

---

🔷 Step 6 — JEE Trap Alert 🚨

❌ Negative area ko subtract kar dena

❌ Distance aur displacement confuse kar lena

❌ Point S ka exact time ignore kar dena

Remember:

For distance:

$$\text{Take modulus of every area}$$

---

✅ Final Answer

$$\boxed{\frac{37}{3}\text{ m}}$$

---

📚 Related Topics

• Total Distance with Acceleration and Deceleration Phases
• Balloon Motion and Falling Object Height Problem
• Water Droplet Spacing Problem Using Kinematics