A 4 kg mass moves under the influence of a force where t is the time in seconds. If the mass starts from the origin at t = 0 the velocity and position after t = 2s will be.

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Solution

Q.
A 4 kg mass moves under the influence of a force $\vec{F} = (4 t^{3} \hat{i} - 3 t \hat{j}) N$ where t is the time in seconds. If the mass starts from the origin at t = 0 the velocity and position after t = 2s will be. [2026]

1 $\vec{v} = 4 \hat{i} - \frac{3}{2} \hat{j} \vec{r} = \frac{8}{5} \hat{i} - \hat{j}$

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

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

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

Ans.
(1)

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

$\vec{a} = \frac{\vec{F}}{m} = t^{3} \hat{i} - \frac{3}{4} t \hat{j}$

$a_{x} = t^{3}$    $a_{y} = - \frac{3}{4} t$

$\frac{d v_{x}}{d t} = t^{3}$    $\frac{d v_{y}}{d t} = - \frac{3}{4}$

$\int_{v_{x} = 0}^{v_{x_{2}}} d v_{x} = \int_{t = 0}^{t = 2} t^{3} d t$ $\int_{0}^{v_{y_{2}}} d v_{y} = \int_{0}^{2} - \frac{3}{4} t d t$

$v_{x_{2}} - 0 = {[\frac{t^{4}}{4}]}_{0}^{2}$ $v_{y_{2}} = - \frac{3}{4} {[\frac{t^{2}}{2}]}_{0}^{2}$

$v_{x_{2}} = 4$ $v_{y_{2}} = \frac{- 3}{2}$

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

$v_{x} = \frac{t^{4}}{4}$ $v_{y} = - \frac{3}{8} t^{2}$

$\int_{0}^{x_{2}} d x = \int_{0}^{2} \frac{t^{4}}{4} d t$ $\int_{0}^{y_{2}} d y = \int_{0}^{2} \frac{- 3}{8} t^{2} d t$

$x_{2} - 0 = {[\frac{t^{5}}{20}]}_{0}^{2}$    $y_{2} - 0 = \frac{- 3}{8} {[\frac{t^{3}}{3}]}_{0}^{2}$

$x_{2} = \frac{8}{5}$    $y_{2} = - 1$

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

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