Question

a particle starting from rest undergoes acceleration given a = |t-2| m/s2. find the vel of particle after 4s​

Answer 1

Answer:

sleep ok

Explanation:

Answer 2

$\small{\fcolorbox{red}{indigo}{\small{\fcolorbox{red}{violet}{\small{\fcolorbox{red}{pink} {\small{\fcolorbox{red}{red}{\small{\fcolorbox{red}{springgreen}{\small{\fcolorbox{red}{blue}{\small{\fcolorbox{red}{yellow} {\small{\fcolorbox{red}{azure} {\small{\fcolorbox{red}{blue}{\small{\fcolorbox{red}{orange}{\huge{\fcolorbox{red}{green}{\large{\fcolorbox{blue}{red}{{\fcolorbox{orange}{aqua}{ANSWER}}}}}}}}}}}}}}}}}}}}}}}}}}$

$\green{\tt{\therefore{Velocity=10.34\:m/s}}}$

$\green{\tt{\therefore{Distance=17.1\:m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Fiven :}} \\ \tt: \implies a = 2 \sqrt{t} \\ \\ \tt: \implies t = 4 \: sec \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Velocity_{ (at \: t = 4 \: sec) } = ? \\ \\ \tt: \implies Displacement_{ (at \: t = 4 \: sec) } =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies a = 2 \sqrt{t} \\ \\ \tt: \implies \frac{dv}{dt} = 2t^{ \frac{1}{2} } \\ \\ \tt: \implies dv = 2 {t}^{ \frac{1}{2} } \times dt \\ \\ \text{Both \: side \: integrating} \\ \\ \tt: \implies \int dv = \int 2 {t}^{ \frac{1}{2} } \times dt \\ \\ \tt: \implies \int dv = 2 \int {t}^{ \frac{1}{2} } \times dt \\ \\ \tt: \implies \int \limits_{0}^{v} dv = 2 \int \limits _{0}^{ 4} \frac{ {t}^{ \frac{1}{2} + 1 } }{ \frac{3}{2}} \\ \\ \tt: \implies v =2 \int \limits _{0}^{ 4} \frac{2t^\frac{3}{2}}{3}\\ \\ \tt: \implies v = 2 \times 2 \times \frac{4^{\frac{3}{2}}}{3}$

$\tt: \implies v = 4 \times \frac{8}{3} \\ \\ \green{\tt: \implies v = \frac{32}{3} = 10.67 \: m/s}$

$\bold{For\:Displacement}\\ \tt : \implies v = \frac{4t^\frac{3}{2} }{3} \\ \\ \tt : \implies \frac{dx}{dt} = \frac{4 {t}^{ \frac{3}{2} } }{3} \\ \\ \tt : \implies dx = \frac{4 {t}^{ \frac{3}{2} } }{3} \times dt \\ \\ \text{Integrating \: both \: side} \\ \\ \tt: \implies \int dx= \int \frac{4 {t}^{ \frac{3}{2} } }{3} \times dt \\ \\ \tt: \implies \int dv = \frac{4}{3} \int {t}^{ \frac{3}{2} } \times dt \\ \\ \tt: \implies \int \limits_{0}^{x} dx = \frac{4}{3} \int \limits _{0}^{ 4} \frac{ {t}^{ \frac{3}{2} + 1 } }{ \frac{3}{2} + 1} \\ \\ \tt: \implies x = \frac{4}{3} \int \limits _{0}^{ 4} \frac{2t^\frac{5}{2}}{5}\\ \\ \tt: \implies x = \frac{4}{3} \times 2\times \frac{4^{\frac{5}{2}}}{5} \\ \\ \tt: \implies x = \frac{8}{3} \times \frac{32}{5} \\ \\ \tt: \implies x = \frac{256}{15} \\ \\ \green{\tt: \implies x = 17.1 \: m}$