A particle starts from the origin at time t = 0 and moves along the positive X-axis. The graph of ve

Q: Why do we calculate the area instead of using the equations of motion like $s = ut + \frac{1}{2}at^2$?

The standard equations of motion are only valid when acceleration is constant. In this graph, the slope (which represents acceleration) changes multiple times. Calculating the area under the $v-t$ graph is a universal method that works regardless of whether acceleration is constant or changing.

Q: What does the vertical line at $t = 4\text{ s}$ physically mean?

A perfectly vertical line on a $v-t$ graph implies an instantaneous change in velocity. In classical mechanics, this would require an infinite acceleration or an infinite force (an impulse) acting for zero time. While physically impossible in the strictest sense, it's often used in problems to model a very sharp, sudden impact.

Q: Why didn't we calculate the area all the way to $t = 8\text{ s}$?

The question specifically asks for the position of the particle at $t = 5\text{ s}$. Therefore, we only care about the displacement that occurred during the first 5 seconds. Any motion that happens after $t = 5\text{ s}$ is irrelevant to this specific question.

Q: What if the graph went below the time axis?

If the velocity-time graph dips below the horizontal time axis, it means the velocity is negative, indicating the particle is moving in the opposite direction. The area bounded by the curve below the axis would be considered negative displacement.

Q: Does the initial position matter when finding the final position from a $v-t$ graph?

Absolutely! The area under the $v-t$ graph only gives you the *change* in position (displacement). To find the final position, you must always use the formula: Final Position = Initial Position + Displacement. In this problem, we were lucky because the initial position was the origin ($0$), so the final position was equal to the displacement.

Question

A particle starts from the origin at time $t = 0$ and moves along the positive X-axis. The graph of velocity with respect to time is shown in figure. What is the position of the particle at time $t = 5	ext{ s}$?

Options:

(A) 6 m

(B) 3 m

(C) 10 m

(D) 9 m

Sigma Percentile · Accepted Answer

### Decoding the Velocity-Time Graph

When you look at a velocity-time ($v-t$) graph, you are essentially looking at the complete biography of a particle's motion. The slope tells you about its acceleration, and the area under the curve reveals its displacement.

In this problem, we are asked to find the position of a particle at $t = 5	ext{ s}$. The problem states that the particle starts from the origin ($x_0 = 0$). This is a crucial piece of information because the final position is given by the initial position plus the displacement. Since the initial position is zero, the final position is exactly equal to the displacement.

### The Power of Area

To find the displacement, we need to calculate the area under the $v-t$ graph from $t = 0$ to $t = 5	ext{ s}$. The mathematical relationship is:

$$x = \int_{0}^{5} v \, dt$$

Instead of dealing with complex integration, we can use simple geometry. The area under the curve up to $t = 5	ext{ s}$ can be broken down into three distinct, easy-to-calculate geometric shapes.

### Breaking Down the Geometry

**Part 1: The Triangle ($0$ to $2	ext{ s}$)**
From $t = 0$ to $t = 2	ext{ s}$, the velocity increases linearly from $0$ to $2	ext{ m/s}$. This forms a right-angled triangle.
$$A_1 = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$$
$$A_1 = \frac{1}{2} 	imes 2 	imes 2 = 2	ext{ m}$$

**Part 2: The First Rectangle ($2$ to $4	ext{ s}$)**
From $t = 2$ to $t = 4	ext{ s}$, the velocity remains constant at $2	ext{ m/s}$. This forms a rectangle.
$$A_2 = 	ext{width} 	imes 	ext{height}$$
$$A_2 = (4 - 2) 	imes 2 = 2 	imes 2 = 4	ext{ m}$$

**Part 3: The Second Rectangle ($4$ to $5	ext{ s}$)**
At exactly $t = 4	ext{ s}$, there is a sudden jump in velocity from $2	ext{ m/s}$ to $3	ext{ m/s}$. From $t = 4$ to $t = 5	ext{ s}$, the velocity is constant at $3	ext{ m/s}$. This forms another rectangle.
$$A_3 = 	ext{width} 	imes 	ext{height}$$
$$A_3 = (5 - 4) 	imes 3 = 1 	imes 3 = 3	ext{ m}$$

### The Final Calculation

Now, we simply sum up these individual areas to find the total displacement.

$$x = A_1 + A_2 + A_3$$
$$x = 2 + 4 + 3 = 9	ext{ m}$$

Since the particle started at the origin, its position at $t = 5	ext{ s}$ is **$9	ext{ m}$**. By breaking the problem down into manageable geometric chunks, we bypassed the need for complex calculus and arrived at the solution elegantly.

Animated Solution for Physics - Kinematics: A particle starts from the origin at time $t = 0$ and moves along the positive X-axis. The graph of velocity with respect to time is shown in figure. What is the position of the particle at time $t = 5 s$ ?

Select Answer:

Visualized Solution

\text{Analyzing the } v-t \text{ Graph}

\text{Concept: Area Under } v-t \text{ Curve}

\text{Splitting the Area}

\text{Calculating } A_1

\text{Calculating } A_2

\text{Calculating } A_3

\text{Total Displacement}

\text{Conclusion}

The Sigma Insight: Motion Graphs

Decoding the Velocity-Time Graph

The Power of Area

Breaking Down the Geometry

The Final Calculation

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