Question 44: Laws of Motion

Question

A 4 kg mass moves under the influence of a force $\vec{F}=\left(4 t^3 \hat{i}-3 t \hat{j}\right) \mathrm{N}$ where $t$ is the time in second. If mass starts from origin at $t=0$, the velocity and position after $t=2 \mathrm{~s}$ will be:

Accepted Answer

According to Newton's Second Law $\overrightarrow{\mathrm{F}}=\mathrm{m} \overrightarrow{\mathrm{a}}$.
$$ \overrightarrow{\mathrm{a}}=\frac{\overrightarrow{\mathrm{F}}}{\mathrm{~m}} $$
Given $\mathrm{m}=4 \mathrm{~kg}$ and $\overrightarrow{\mathrm{F}}=4 \mathrm{t}^3 \hat{\imath}-3 \mathrm{t} \hat{\jmath}$ :
$$ \vec{a}=\frac{4 t^3 \hat{\imath}-3 t \hat{\jmath}}{4}=\left(t^3 \hat{\imath}-\frac{3}{4} t \hat{\jmath}\right) \mathrm{m} / \mathrm{s}^2 $$
Acceleration is the rate of change of velocity :
$$ \vec{a}=\frac{d \vec{v}}{d t} $$
Since the mass starts from rest at the origin at $t=0$, we have $\vec{v}(0)=0$
$$ \int\limits_0^{\vec{v}} d \vec{v}=\int\limits_0^t\left(t^3 \hat{\imath}-\frac{3}{4} t \hat{\jmath}\right) d t $$
$$\Rightarrow $$ $$ \overrightarrow{\mathrm{v}}=\left(\frac{\mathrm{t}^4}{4} \hat{\imath}-\frac{3 \mathrm{t}^2}{8} \hat{\jmath}\right) \mathrm{m} / \mathrm{s} $$
So, velocity at $\mathrm{t}=2 \mathrm{~s}$ is :
$$ \overrightarrow{\mathrm{v}}=\frac{2^4}{4} \hat{\imath}-\frac{3\left(2^2\right)}{8} \hat{\jmath}=\frac{16}{4} \hat{\imath}-\frac{12}{8} \hat{\jmath} $$
$$\Rightarrow $$ $$ \vec{v}=\left(4 \hat{\imath}-\frac{3}{2} \hat{\jmath}\right) \mathrm{m} / \mathrm{s} $$
Velocity is the rate of change of position :
$$ \vec{v}=\frac{d \vec{r}}{d t} \Rightarrow d \vec{r}=\vec{v} d t $$
At $\mathrm{t}=0$, the mass is at the origin ( $\overrightarrow{\mathrm{r}}=0$ ),
$$ \int\limits_0^{\overrightarrow{\mathrm{r}}} \mathrm{~d} \overrightarrow{\mathrm{r}}=\int\limits_0^{\mathrm{t}}\left(\frac{\mathrm{t}^4}{4} \hat{\imath}-\frac{3 \mathrm{t}^2}{8} \hat{\jmath}\right) \mathrm{dt} $$
$$\Rightarrow $$ $\overrightarrow{\mathrm{r}}=\left(\frac{\mathrm{t}^5}{20} \hat{\imath}-\frac{\mathrm{t}^3}{8} \hat{\jmath}\right)_0^{\mathrm{t}} \mathrm{m}$
$$\Rightarrow $$ $\overrightarrow{\mathrm{r}}=\left[\frac{\mathrm{t}^5}{20} \hat{\imath}-\frac{\mathrm{t}^3}{8} \hat{\jmath}\right]-\left[\frac{0^5}{20} \hat{\imath}-\frac{0^3}{8} \hat{\jmath}\right] \mathrm{m}$
$$\Rightarrow $$ $$ \vec{r}=\left(\frac{t^5}{20} \hat{\imath}-\frac{t^3}{8} \hat{\jmath}\right) $$
Now, calculate position at $\mathrm{t}=2 \mathrm{~s}$ :
$$ \overrightarrow{\mathrm{r}}=\frac{2^5}{20} \hat{\imath}-\frac{2^3}{8} \hat{\jmath}=\frac{32}{20} \hat{\imath}-\frac{8}{8} \hat{\jmath} $$
$$\Rightarrow $$ $\overrightarrow{\mathrm{r}}=\frac{8}{5} \hat{\imath}-1 \hat{\jmath}=\left(\frac{8}{5} \hat{\imath}-\hat{\jmath}\right) \mathrm{m}$
Therefore, at $t=2 \mathrm{~s}$, the velocity is $\vec{v}=4 \hat{\imath}-\frac{3}{2} \hat{\jmath}$ and the position vector is $\vec{r}=\frac{8}{5} \hat{\imath}-\hat{\jmath}$.
Hence, the correct option is (C).

Answer

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

Answer

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

Answer

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

Answer

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

Laws of Motion

A 4 kg mass moves under the influence of a force \vec{F}=\left(4 t^3 \hat{i}-3 t \hat{j}\right) \mathrm{N} where t is the time in second. If mass starts from origin at t=0, the velocity and position after t=2 \mathrm{~s} will be:

Choose the correct answer: