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Velocity of a particle moving in a straight line varies with its displacement as...

Question:

Velocity of a particle moving in a straight line varies with its displacement as {eq}v=(\sqrt{4+4 \ s})m/s {/eq}. Displacement of particle at time {eq}t=0 {/eq} is {eq}s=0 {/eq}. Find displacement of particle at time {eq}t=2 \ s. {/eq}

The velocity:

The rate of change of the displacement with time is called the velocity. It is a vector quantity that has the magnitude as well as the direction. The SI unit of the velocity is {eq}\text{m/s} {/eq}.

The velocity of an object is formulated as:

$$\begin{align} \color{red}{\vec{v}=\frac{\vec{s}}{t}} \end{align} $$

Here:

  • {eq}\vec{s} {/eq} is the displacement vector.
  • {eq}t {/eq} is the time period.

Answer and Explanation:


Given:

  • The velocity of the particle in the form of displacement is {eq}\begin{align} \Rightarrow v=\sqrt{4+4 \ s} \end{align} {/eq}


Now to find the displacement in the form of temperature:

{eq}\begin{align} \Rightarrow \frac{\mathrm{d} s}{\mathrm{d} t}=\sqrt{4+4 \ s} \end{align} {/eq}


{eq}\begin{align} \Rightarrow \int_{0}^{s}\frac{ds}{\sqrt{4+4 \ s}}=\int_{0}^{2}dt\\ \end{align} {/eq}


{eq}\begin{align} \Rightarrow \left[2(\sqrt{4+4 \ s})\right]_{0}^{s}=\left[t\right]_{0}^{2} \end{align} {/eq}


{eq}\begin{align} \Rightarrow 2\left(\sqrt{4-4s}-\sqrt{4}\right)=2 \end{align} {/eq}


{eq}\begin{align} \Rightarrow \sqrt{4-4s}-2=1 \end{align} {/eq}

{eq}\begin{align} \Rightarrow \left(\sqrt{4+4 \ s}\right)^2=(3)^2 \end{align} {/eq}


Thus the displacement of the particle is:

{eq}\begin{align} \Rightarrow s&=\frac{(9-4)}{4} \ \text{m}\\ &=\color{blue}{\frac{5}{4} \ \text{m}} \end{align} {/eq}


Learn more about this topic:

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Linear Velocity: Definition & Formula

from MCAT Prep: Help and Review

Chapter 14 / Lesson 12
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