Q69P - University Physics with Modern Physics

Question

The acceleration of a particle is given by ${\mathbf{a}}_x\left(\mathbf{t}\right)=-\mathbf{2}.\mathbf{00}\mathbf{m}/{\mathbf{s}}^2+(\mathbf{3}.\mathbf{00}\mathbf{m}/{\mathbf{s}}^3)\mathbf{t}$ . (a) Find the initial velocity  ${\mathbf{v}}_{0x}$ such that the particle will have the same x-coordinate at $\mathbf{t}=\mathbf{4}.\mathbf{00}\mathbf{s}$ as it had at  $\mathbf{t}\mathbf{=}\mathbf{0}$. (b) What will be the velocity at  $\mathbf{t}=\mathbf{4}.\mathbf{00}\mathbf{s}$?

Step-by-Step Solution

Answer

a) The initial velocity of the particle is  $-4\text{\hspace{0.17em}m/s}$.

b) The velocity at is $12\text{\hspace{0.17em}m/s}$ .

Step 1: Identification of the given data

The given data can be listed below as,

• The particle will have the same x-coordinate at  $\mathrm{t}=4.00\text{\hspace{0.17em}s}$.
• The particle will have the same x-coordinate at  $\mathrm{t}=0$.

Step 2: Significance of the Newton’s first law to evaluate the initial and the final velocity

This law states that a particular object will continue to move in a uniform motion unless it is resisted by an external force.

Double integrating the equation of acceleration gives the equation of the displacement that gives the initial velocity of the particle. Moreover, the equation of velocity also gives the velocity of the particle.

Step 3: Determination of initial velocity of the final velocity

The given equation can be expressed as:

${\mathrm{a}}_{\mathrm{x}}\left(\mathrm{t}\right)=-2.00{\text{\hspace{0.17em}m/s}}^{\text{2}}\text{+}\left({\text{3.00\hspace{0.17em}m/s}}^{\text{3}}\right)\mathrm{t}$   … i)

We know that, $\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}$

Hence, using this, the equation i) can be written as:

$\begin{array}{c}\int \mathrm{dv}=\int \mathrm{a}.\mathrm{dt}\\ \mathrm{v}=\int (-2+3\mathrm{t})\mathrm{dt}\\ \mathrm{v}=-2\mathrm{t}+\frac{3{\mathrm{t}}^2}{2}+{\mathrm{v}}_0\\ \mathrm{v}=-2\mathrm{t}+1.5{\mathrm{t}}^2+{\mathrm{v}}_0\end{array}$

Here,${\mathrm{v}}_0$ is the initial velocity.

We also know that, $\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}$

Hence, using this, the equation i) can be written as:

$\begin{array}{c}\int \mathrm{ds}=\int \mathrm{v}.\mathrm{dt}\\ \mathrm{s}=\int (-2\mathrm{t}+1.5{\mathrm{t}}^2+{\mathrm{v}}_0)\mathrm{dt}\\ \mathrm{s}=\frac{-2{\mathrm{t}}^2}{2}+\frac{1.5{\mathrm{t}}^3}{3}+{\mathrm{v}}_0\mathrm{t}+{\mathrm{x}}_0\\ \mathrm{s}=-{\mathrm{t}}^2+0.5{\mathrm{t}}^3+{\mathrm{v}}_0\mathrm{t}+{\mathrm{x}}_0\end{array}$

Here, ${\mathrm{x}}_0$ is the initial displacement.

a) 

According to the question, the equation of displacement at $\mathrm{t}=4.00\text{\hspace{0.17em}s}$ and $\mathrm{t}=0\text{\hspace{0.17em}s}$ is equal.

Hence, the equation of displacement at $\mathrm{t}=0\text{\hspace{0.17em}s}$ is-

${\mathrm{s}}_{\mathrm{at}\mathrm{t}=0}={\mathrm{x}}_0$… ii)

Hence, the equation of displacement at is-

${\mathrm{s}}_{\mathrm{at}\mathrm{t}=4.00\mathrm{s}}=16+4{\mathrm{v}}_0+{\mathrm{x}}_0$… iii)

Equating the equation ii) and iii), we get-

$\begin{array}{l}{\mathrm{x}}_0=16+4{\mathrm{v}}_0+{\mathrm{x}}_0\\ {\mathrm{v}}_0=-4\text{\hspace{0.17em}m/s}\end{array}$

Thus, the initial velocity of the particle is $-4\text{\hspace{0.17em}m/s}$.

b)

Using the equation of velocity, the velocity at $\mathrm{t}=4.00\text{\hspace{0.17em}s}$ is expressed as:

$\begin{array}{c}\mathrm{v}=-2\mathrm{t}+1.5{\mathrm{t}}^2+{\mathrm{v}}_0\\ \mathrm{v}=-8+24-4\\ \mathrm{v}=12\text{\hspace{0.17em}m/s}\end{array}$

Thus, the velocity at  $\mathrm{t}=4.00\text{\hspace{0.17em}s}$ is  $12\text{\hspace{0.17em}m/s}$ .