The-initial-velocity-of-a-particle-is-u-at-t-0-and-acceleration-f-is-given-by-f-at-where-t-is-time-and-a-is-a-constant-Which-of-the-following-relations-is-valid-1-v-u-at-2-2-v-u-

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Question Number 13478 by Tinkutara last updated on 20/May/17

$$\mathrm{The}\mathrm{initial}\mathrm{velocity}\mathrm{of}\mathrm{a}\mathrm{particle}\mathrm{is}u$$$$(\mathrm{at}t=0)\mathrm{and}\mathrm{acceleration}f\mathrm{is}\mathrm{given}\mathrm{by}$$$$f=at\mathrm{where}t\mathrm{is}\mathrm{time}\mathrm{and}{}^{\prime }{a}^{\prime }\mathrm{is}\mathrm{a}\mathrm{constant}.$$$$\mathrm{Which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{relations}\mathrm{is}$$$$\mathrm{valid}?$$$$\left(1\right)v=u+{at}^2$$$$\left(2\right)v=u+\frac{1}{2}{at}^2$$$$\left(3\right)v=u+at$$$$\left(4\right)v=u$$

Answered by ajfour last updated on 20/May/17

$$\frac{dv}{dt}=f=at$$$$\Rightarrow {\int }_u^{v}dv={\int }_0^{t}atdt$$$$v-u=\frac{1}{2}{at}^2$$$$v=u+\frac{1}{2}{at}^2.$$$$\left(2\right)isvalid.$$

Commented by ajfour last updated on 20/May/17

$$isitfinenow?$$

Commented by Tinkutara last updated on 20/May/17

$$\mathrm{OK}.\mathrm{Thanks}!$$

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