Test Index

Laws of Motion Part 2

Section: Physics

Question : 12 of 52

Marks: +1, -0

\vec{F}= (4 t^3 \hat{i}-3 t \hat{j}) \mathrm{N}

t

t=0

t=2 \mathrm{s}

$\vec{v}=3 \hat{i}+\frac{3}{2} \hat{j} \quad \vec{r}=\frac{6}{5} \hat{i}+\hat{j}$
$\vec{v}=4 \hat{i}-\frac{3}{2} \hat{j} \quad \vec{r}=\frac{6}{5} \hat{i}-\hat{j}$
$\vec{v}=4 \hat{i}-\frac{3}{2} \hat{j} \quad \vec{r}=\frac{8}{5} \hat{i}-\hat{j}$
$\vec{v}=4 \hat{i}+\frac{5}{2} \hat{j} \quad \vec{r}=\frac{8}{5} \hat{i}+2 \hat{j}$

Solution:

Concept:

Newton's second law gives acceleration from force.

Integrate acceleration to get velocity, then integrate velocity to get position.

Explanation:

Given

\mathrm{m}=4\ \mathrm{kg}

and

\vec{F}=4t^3\hat{i}-3t\hat{j}\ \mathrm{N}

Acceleration

\vec{a}=\frac{\vec{F}}{m}=t^3\hat{i}-\frac{3}{4}t\hat{j}\ \mathrm{m/s}^2

Velocity

\vec{v}=\int_0^t \vec{a}\,dt

with

\vec{v}(0)=0

\vec{v}=\int_0^t (t^3\hat{i}-\frac{3}{4}t\hat{j})dt = \frac{t^4}{4}\hat{i}-\frac{3t^2}{8}\hat{j}\ \mathrm{m/s}

t=2\ \mathrm{s}

\vec{v}=\frac{16}{4}\hat{i}-\frac{12}{8}\hat{j}=4\hat{i}-\frac{3}{2}\hat{j}\ \mathrm{m/s}

Position

\vec{r}=\int_0^t \vec{v}\,dt

with

\vec{r}(0)=0

\vec{r}=\int_0^t (\frac{t^4}{4}\hat{i}-\frac{3t^2}{8}\hat{j})dt = \frac{t^5}{20}\hat{i}-\frac{t^3}{8}\hat{j}\ \mathrm{m}

t=2\ \mathrm{s}

\vec{r}=\frac{32}{20}\hat{i}-\frac{8}{8}\hat{j}=\frac{8}{5}\hat{i}-\hat{j}\ \mathrm{m}

Answer:

\vec{v}=4\hat{i}-\frac{3}{2}\hat{j}

and

\vec{r}=\frac{8}{5}\hat{i}-\hat{j}

. Option (C).

Go to Question:

Prev Question

Next Question