Question and Answers Forum

All Questions      Topic List

Vector Questions

Previous in All Question      Next in All Question      

Previous in Vector      Next in Vector      

Question Number 126475 by sdfg last updated on 20/Dec/20

Answered by physicstutes last updated on 20/Dec/20

$$\left(\mathrm{a}\right)\mathbf{a}=\frac{\mathbf{\Delta }\mathbf{v}}{\mathbf{\Delta }t}=\frac{[\left(10\mathbf{j}\right)-(10\mathbf{i}+20\mathbf{j})]\mathrm{m}{\mathrm{s}}^{-1}}{(4-0)\mathrm{s}}=[-\left(\frac{10}{4}\right)\mathbf{i}-\left(\frac{10}{4}\right)\mathbf{j}]\mathrm{m}{\mathrm{s}}^{-2}$$$$\mathrm{or}\mathbf{a}=(-\frac{5}{2}\mathbf{i}-\frac{5}{2}\mathbf{j})\mathrm{m}{\mathrm{s}}^{-2}\mathrm{ofcourse}\mathrm{assuming}\mathrm{we}\mathrm{are}\mathrm{using}\mathrm{SI}\mathrm{base}\mathrm{units}.$$$$\left(\mathrm{b}\right)\mathbf{r}={\mathbf{r}}_0+{\mathbf{v}}_{0}t+\frac{1}{2}\mathbf{a}{t}^2\mathrm{but}{\mathbf{r}}_0=\overrightarrow{0}\mathrm{since}\mathrm{it}\mathrm{is}\mathrm{initially}\mathrm{at}\mathrm{the}\mathrm{origin}.$$$$\Rightarrow \mathbf{r}=(10\mathbf{i}+20\mathbf{j})t+\left[(-\frac{5}{4}\mathbf{i}-\frac{5}{4}\mathbf{j})\right]t^2$$$$\mathbf{v}={\mathbf{v}}_0+\mathbf{a}t$$$$\Rightarrow \mathbf{v}=(10\mathbf{i}+20\mathbf{j})+(-\frac{5}{2}\mathbf{i}-\frac{5}{2}\mathbf{j})t$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com