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Question Number 42357 by preet last updated on 24/Aug/18

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Aug/18

v^→ =(dr^→ /dt)=(d/dt)(3ti−t^2 j+4k)  v^→ =3i−2tj+0.k  (v^→ )_(t=5sec) =3i−10j+0.k  so ∣(v^→ )_(t=5sec) ∣=(√(3^2 +(−10)^2 +0^2 )) =(√(109))  meter/sec

$$\overset{\rightarrow} {{v}}=\frac{{d}\overset{\rightarrow} {{r}}}{{dt}}=\frac{{d}}{{dt}}\left(\mathrm{3}{ti}−{t}^{\mathrm{2}} {j}+\mathrm{4}{k}\right) \\ $$$$\overset{\rightarrow} {{v}}=\mathrm{3}{i}−\mathrm{2}{tj}+\mathrm{0}.{k} \\ $$$$\left(\overset{\rightarrow} {{v}}\right)_{{t}=\mathrm{5}{sec}} =\mathrm{3}{i}−\mathrm{10}{j}+\mathrm{0}.{k} \\ $$$${so}\:\mid\left(\overset{\rightarrow} {{v}}\right)_{{t}=\mathrm{5}{sec}} \mid=\sqrt{\mathrm{3}^{\mathrm{2}} +\left(−\mathrm{10}\right)^{\mathrm{2}} +\mathrm{0}^{\mathrm{2}} }\:=\sqrt{\mathrm{109}}\:\:{meter}/{sec} \\ $$

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