An object, moving with a speed of $$6.25 \, \text{m/s}$$, 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:

Solution & Explanation

1. Set Up the Differential Equation

The problem provides the deceleration rate of the object as a function of its instantaneous speed ($$v$$):

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

To solve for time ($$t$$), we rearrange the equation using the separation of variables method, grouping all terms involving velocity ($$v$$) on one side and time ($$t$$) on the other side:

$$\frac{1}{\sqrt{v}} \cdot dv = -2.5 \cdot dt$$

$$v^{-\frac{1}{2}} \cdot dv = -2.5 \cdot dt$$

2. Integrate with Initial and Final Limits

We apply definite integration on both sides matching the physical constraints of the motion:

• At initial time $$t = 0 \,\, \text{s}$$, the initial speed is $$v = 6.25 \,\, \text{m/s}$$.
• At final time $$t = t$$, the object comes to rest, meaning the final speed is $$v = 0 \,\, \text{m/s}$$.

Setting up the integration boundaries:

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

Using the power rule for integration ($$\int v^n \cdot dv = \frac{v^{n+1}}{n+1}$$):

$$\left[ \frac{v^{\frac{1}{2}}}{\frac{1}{2}} \right]_{6.25}^{0} = -2.5 \cdot [t]_{0}^{t}$$

$$2 \cdot \left[ \sqrt{v} \right]_{6.25}^{0} = -2.5 \cdot (t - 0)$$

3. Evaluate Limits and Compute Time ($$t$$)

Substitute the upper and lower boundary values into the equation:

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

Since $$\sqrt{6.25} = 2.5$$:

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

$$-5 = -2.5 \cdot t$$

Isolating the time variable ($$t$$):

$$t = \frac{-5}{-2.5} = 2 \,\, \text{s}$$

Concept Check: Because the deceleration rate is non-linear and depends directly on $$\sqrt{v}$$, the slowing force reduces continuously as the speed drops, resulting in a smooth stop in exactly 2 seconds.

Correct Option Key: Option A ($2 \,\, \text{s}$)