A particle starts from rest. Its acceleration (a) versus time (t) is as shown in the figure. The max

Q: Wait, why does the maximum velocity occur at $t = 11 \text{ s}$ when the acceleration is zero there? Shouldn't velocity be zero if acceleration is zero?

This is a very common trap! Acceleration is the *rate of change* of velocity, not the velocity itself. A positive acceleration means velocity is increasing. From $t = 0$ to $t = 11$, acceleration is positive (even though it's decreasing), so velocity is constantly going up. At $t = 11$, acceleration becomes zero, meaning the velocity *stops increasing*. Therefore, it has reached its peak value at that exact moment.

Q: Why can't we just use the kinematic equation $v = u + at$?

The equation $v = u + at$ is strictly valid only when the acceleration $a$ is **constant**. In this graph, the acceleration is changing continuously (it's a downward sloping line, not a horizontal one). When acceleration is variable, you must use calculus or graphical methods (like finding the area) to find the velocity.

Q: What if the graph continued below the time axis after $t = 11 \text{ s}$?

If the graph dipped below the time axis, the acceleration would become negative (retardation). This means the particle would start slowing down. The area under the time axis would be considered negative, and you would subtract it from the positive area to find the net change in velocity. The maximum velocity would still be at $t = 11 \text{ s}$ right before it starts slowing down.

Q: How do we know the initial velocity is zero? What if it wasn't?

The problem explicitly states, "A particle starts from rest." This is the physics code-word for $v_i = 0$. If it had an initial velocity, say $u = 5 \text{ m/s}$, the area under the graph would still give you the *change* in velocity ($\Delta v = 55$). You would then find the final velocity by adding the initial velocity: $v_f = u + \Delta v = 5 + 55 = 60 \text{ m/s}$.

Q: Is the slope of this graph useful for anything?

Yes! The slope of an acceleration-time graph represents the rate of change of acceleration, which is a quantity known as "jerk" or "jolt". In this specific problem, the slope is constant and negative ($\frac{-10}{11} \text{ m/s}^3$), meaning the particle is experiencing a constant negative jerk. However, we don't need the jerk to find the maximum velocity.

Question

A particle starts from rest. Its acceleration ($a$) versus time ($t$) is as shown in the figure. The maximum speed of the particle will be

Options:

(A) $110 	ext{ m/s}$

(B) $55 	ext{ m/s}$

(C) $550 	ext{ m/s}$

(D) $660 	ext{ m/s}$

Sigma Percentile · Accepted Answer

Title: The Geometry of Motion: Decoding the Acceleration-Time Graph

### Analyzing the Setup

Imagine you are sitting in a sports car that is just about to launch. At the very first instant, the moment you press the pedal, you feel the maximum push—the maximum acceleration. But as the car speeds up, that pushing force gradually decreases until it completely vanishes. This physical experience is exactly what the given acceleration-time graph represents!

We are presented with a straight-line graph sloping downwards. At time $t = 0$, the acceleration is at its peak value of $10 	ext{ m/s}^2$. As time ticks forward, this acceleration linearly drops, finally hitting $0 	ext{ m/s}^2$ at exactly $t = 11 	ext{ s}$.

The problem also hands us a crucial piece of initial data: "A particle starts from rest." In the language of kinematics, this translates to an initial velocity of zero.

$$v_i = 0 	ext{ m/s}$$

Our mission is to find the maximum speed the particle achieves during this 11-second interval.

### The Master Equation

To solve this, we need to build a bridge between the graph we have (acceleration vs. time) and the quantity we want (velocity). Let's return to the fundamental definition of acceleration. Acceleration is the rate at which velocity changes with respect to time.

$$a = \frac{dv}{dt}$$

If we rearrange this differential equation to isolate the change in velocity, we get:

$$dv = a \cdot dt$$

To find the total change in velocity over a time interval, we must integrate both sides of this equation:

$$\int_{v_i}^{v_f} dv = \int_{t_i}^{t_f} a \cdot dt$$

Here is the beautiful part: in calculus, the definite integral of a function represents the area under its curve. Therefore, the integral of acceleration with respect to time is simply the geometric area under the $a-t$ graph!

$$\Delta v = 	ext{Area under } a-t 	ext{ graph}$$

It is also important to note **when** the maximum velocity occurs. Since the acceleration is positive for the entire duration from $t = 0$ to $t = 11 	ext{ s}$, the velocity is continuously increasing. The moment the acceleration hits zero at $t = 11 	ext{ s}$, the velocity stops increasing. Thus, the velocity is at its absolute maximum right at the end of this triangle.

### Final Calculation

Now, let's look at the shape bounded by our graph and the coordinate axes. It forms a perfect right-angled triangle. We can easily calculate its area using the standard geometric formula:

$$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$$

Looking at the axes, the base of our triangle lies along the time axis, stretching from $0$ to $11 	ext{ s}$.

$$	ext{Base} = 11 	ext{ s}$$

The height of the triangle lies along the acceleration axis, starting from $0$ and peaking at $10 	ext{ m/s}^2$.

$$	ext{Height} = 10 	ext{ m/s}^2$$

Let's substitute these values into our area formula:

$$\Delta v = \frac{1}{2} 	imes 11 	imes 10$$

Executing the multiplication:

$$\Delta v = 11 	imes 5 = 55 	ext{ m/s}$$

We have found that the total change in velocity is $55 	ext{ m/s}$. But remember, the change in velocity is the final velocity minus the initial velocity:

$$\Delta v = v_{max} - v_i$$

Since the particle started from rest, $v_i = 0$. Substituting this into our equation:

$$55 = v_{max} - 0$$

Therefore, the maximum speed attained by the particle is:

$$v_{max} = 55 	ext{ m/s}$$

This elegant geometric approach leads us directly to the correct answer, which corresponds to option (b).

Animated Solution for Physics - Kinematics: A particle starts from rest. Its acceleration ( $a$ ) versus time ( $t$ ) is as shown in the figure. The maximum speed of the particle will be

Select Answer:

Visualized Solution

Analyzing the $a - t$ Graph

Extracting Initial Conditions

The Kinematic Relationship

Area Under the Curve

Identifying the Geometry

Setting up the Height

Calculating the Area

Concluding Maximum Speed

The Sigma Insight: Motion Graphs

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

The speed versus time graph for a particle is shown in the figure. The distance travelled (in metre) by the particle during the time interval $t = 0 s$ to $t = 5 s$ will be ……… .

A particle is moving with constant acceleration $a$ . Following graph shows $v^{2}$ versus $x$ (displacement) plot. The acceleration of the particle is ........ $m/s^{2}$

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 $t = 5 s$ ?

Two particles A and B start from the same point and move in the positive $x$ -direction. In a time interval of $2.00 s$ after they start, their velocities $v$ vary with time $t$ as shown in the following figures. What is the maximum separation between the particles during this time interval?

The velocity- displacement graph of a particle is shown in the figure. The acceleration-displacement graph of the same particle is represented by

A material particle is chasing another one and both of them are moving on the same straight line. After they pass a particular point, their velocities $v$ vary with time $t$ as shown in the figure. When will the chase end?

The v-t graph of a body in a straight line motion is shown in the figure. The point S is at 4.333 s. The total distance covered by the body in 6 s is

A particle starts from origin $O$ from rest and moves with a uniform acceleration along the positive X-axis. Identify all figures that correctly represent the motion qualitatively. ( $a = acceleration$ , $v = velocity$ , $x = displacement$ , $t = time$ )

A ball is thrown vertically upwards. Its distance $s$ from a fixed point varies with time $t$ according to the following graph. Calculate velocity of projection of the ball.