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

### Related Formula veca = fracvecFm$$\vec{a} = \frac{\vec{F}}{m}$$ vecv = int veca \, dt$$\vec{v} = \int \vec{a} \, dt$$ vecr = int vecv \, dt$$\vec{r} = \int \vec{v} \, dt$$ ### Core Logic Given vecF = 4t^3hati - 3thatj$\vec{F} = 4t^3\hat{i} - 3t\hat{j}$ and mass m = 4text kg$m = 4\text{ kg}$. Acceleration: veca = fracvecFm = frac4t^3hati - 3thatj4 = t^3hati - frac34thatj$$\vec{a} = \frac{\vec{F}}{m} = \frac{4t^3\hat{i} - 3t\hat{j}}{4} = t^3\hat{i} - \frac{3}{4}t\hat{j}$$ ### Step 1: Calculating Velocity Velocity is the integral of acceleration with v(0) = 0$v(0) = 0$: vecv = int left( t^3hati - frac34thatj right) dt = fract^44hati - frac38t^2hatj$$\vec{v} = \int \left( t^3\hat{i} - \frac{3}{4}t\hat{j} \right) dt = \frac{t^4}{4}\hat{i} - \frac{3}{8}t^2\hat{j}$$ At t = 2text s$t = 2\text{ s}$: vecv(2) = frac(2)^44hati - frac38(2)^2hatj = 4hati - frac32hatj$$\vec{v}(2) = \frac{(2)^4}{4}\hat{i} - \frac{3}{8}(2)^2\hat{j} = 4\hat{i} - \frac{3}{2}\hat{j}$$ ### Step 2: Calculating Position Position is the integral of velocity with vecr(0) = vec0$\vec{r}(0) = \vec{0}$: vecr = int vecv \, dt = int left( fract^44hati - frac38t^2hatj right) dt$$\vec{r} = \int \vec{v} \, dt = \int \left( \frac{t^4}{4}\hat{i} - \frac{3}{8}t^2\hat{j} \right) dt$$ vecr = fract^520hati - frac324t^3hatj = fract^520hati - fract^38hatj$$\vec{r} = \frac{t^5}{20}\hat{i} - \frac{3}{24}t^3\hat{j} = \frac{t^5}{20}\hat{i} - \frac{t^3}{8}\hat{j}$$ At t = 2text s$t = 2\text{ s}$: vecr(2) = frac(2)^520hati - frac(2)^38hatj = frac3220hati - frac88hatj = frac85hati - hatj$$\vec{r}(2) = \frac{(2)^5}{20}\hat{i} - \frac{(2)^3}{8}\hat{j} = \frac{32}{20}\hat{i} - \frac{8}{8}\hat{j} = \frac{8}{5}\hat{i} - \hat{j}$$ ### Pattern Recognition Recognize F to a to v to r$F \to a \to v \to r$ means consecutive integrations. Because the initial state is from rest at origin, we don't have to worry about constants of integration. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion Class 11 Physics: Kinematics