An object moving with a speed of 6.25 m/s, is decelerated at a rate given by dv/dt = -2.5 ∑v, where

Q: Why did we use $\frac{dv}{dt}$ instead of $v \frac{dv}{dx}$ for acceleration?

Since we need to find the *time* taken to come to rest, it is much more direct to use the time derivative of velocity $\frac{dv}{dt}$. If we needed the stopping *distance*, $v \frac{dv}{dx}$ would be the better choice.

Q: How do I quickly find the square root of $6.25$?

You can think of $6.25$ as $\frac{625}{100}$. The square root of $625$ is $25$, and the square root of $100$ is $10$. So, $\frac{25}{10} = 2.5$.

Q: What happens to the negative sign in the acceleration?

The negative sign indicates deceleration (velocity is decreasing). It naturally cancels out with the negative sign that arises when we subtract the initial velocity limit from the final zero velocity limit during integration.

Q: Can I use the standard equation $v = u + at$ here?

No! The standard equations of kinematics are strictly for *constant* acceleration. Here, the acceleration depends on velocity ($a \propto \sqrt{v}$), so it is variable. You must use calculus.

Q: Why is the final velocity limit zero?

The question asks for the time taken for the object to "come to rest". "Rest" physically means the object has stopped moving, so its final velocity $v_f = 0$.

Question

An object moving with a speed of $6.25	ext{ m/s}$, is decelerated at a rate given by $\frac{dv}{dt} = -2.5 \sqrt{v}$, where $v$ is the instantaneous speed. The time taken by the object to come to rest, would be

Options:

(A) $2	ext{ s}$

(B) $4	ext{ s}$

(C) $8	ext{ s}$

(D) $1	ext{ s}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine an object sliding across a rough surface. It starts with a brisk initial speed of $v_0 = 6.25 	ext{ m/s}$. However, it doesn't maintain this speed. It experiences a deceleration, meaning its acceleration vector points in the exact opposite direction of its motion.

What makes this problem interesting is that the deceleration isn't constant. The rate at which it slows down depends on how fast it's currently moving, given by the differential equation $\frac{dv}{dt} = -2.5\sqrt{v}$. Our mission is to find out exactly how long it takes for this object to completely come to rest.

### The Master Equation
To find the stopping time, we need a relationship between velocity ($v$) and time ($t$). We start with our given differential equation:

$$\frac{dv}{dt} = -2.5\sqrt{v}$$

This equation is separable. We can group all the velocity terms on one side and the time terms on the other. By dividing both sides by $\sqrt{v}$ and multiplying by $dt$, we get:

$$\frac{dv}{\sqrt{v}} = -2.5 dt$$

Now, the equation is perfectly primed for integration.

### Setting up the Integral
We must integrate both sides with their respective physical limits. At the very beginning, when time $t = 0$, the initial velocity is $6.25 	ext{ m/s}$. We want to find the time $t$ when the object comes to rest, which means its final velocity is $0 	ext{ m/s}$.

Setting up the definite integral, we write:

$$\int_{6.25}^{0} v^{-1/2} dv = \int_{0}^{t} -2.5 dt$$

Notice how the lower limits ($6.25$ and $0$) correspond to the initial state, and the upper limits ($0$ and $t$) correspond to the final state.

### Executing the Calculus
Let's evaluate the integrals. On the left side, using the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1}$, the integral of $v^{-1/2}$ becomes $\frac{v^{1/2}}{1/2}$, which simplifies to $2\sqrt{v}$. On the right side, the integral of a constant with respect to $dt$ is just the constant multiplied by $t$.

$$2 \left[ \sqrt{v} ight]_{6.25}^{0} = -2.5 \left[ t ight]_{0}^{t}$$

Now, we carefully substitute our upper and lower limits:

$$2 (\sqrt{0} - \sqrt{6.25}) = -2.5 (t - 0)$$

We know that the square root of $6.25$ is exactly $2.5$. Substituting this value in:

$$2 (0 - 2.5) = -2.5 t$$

$$-5 = -2.5 t$$

### Final Calculation
The negative signs on both sides tell a beautiful physical story: the loss of velocity perfectly balances the negative acceleration over time. Solving for $t$:

$$t = \frac{-5}{-2.5}$$

**$$t = 2 	ext{ s}$$**

The object will take exactly $2$ seconds to come to a complete halt. This elegant result shows how calculus seamlessly models dynamic physical systems.

Animated Solution for Physics - Kinematics: An object moving with a speed of $6.25 m/s$ , is decelerated at a rate given by $\frac{d v}{d t} = - 2.5 v$ , where $v$ is the instantaneous speed. The time taken by the object to come to rest, would be

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Visualized Solution

$Initial Setup$

$Differential Equation of Motion$

$Setting up the Integral$

$Integrating both sides$

$Applying Limits$

$Calculating Time$

$Further Analysis$

The Sigma Insight: Motion in a Straight Line

Analyzing the Setup Imagine an object sliding across a rough surface

The Master Equation To find the stopping time, we need a relationship between velocity ( $v$ ) and time ( $t$ )

Setting up the Integral We must integrate both sides with their respective physical limits

Executing the Calculus Let's evaluate the integrals

Final Calculation The negative signs on both sides tell a beautiful physical story: the loss of velocity perfectly balances the negative acceleration over time

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Differential Equation of Motion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up the Integral

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Integrating both sides

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying Limits

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Time

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Further Analysis