Question

a velocity of a particle starting from rest is given by v=2t^2-3t+4. then its displacement at the end of 1 second is?​

Answer 1

Answer:

Given;

• velocity of the particle starting from rest is given as: v = 2t² - 3t + 4

Unknown;

•Displacement of the particle at the end of 1 second.

Solution;

•The velocity of a particle at any instant of time is known as instantaneous velocity. It is given as:

$\longrightarrow\displaystyle\bf v = \lim_{\Delta \to 0} \dfrac{\Delta x}{\Delta t}$

OR,

$\longrightarrow\displaystyle\bf v = \dfrac{dx}{dt}$

•Rearranging the above formula we have:

$\longrightarrow\displaystyle\sf dx = v.dt$

• By Integrating both the sides we have:

$\longrightarrow\displaystyle \sf\int\limits_{0}^{x}dx=\int\limits_{0}^{1} v.dt$

$\longrightarrow\displaystyle \sf\int\limits_{0}^{x}dx=\int\limits_{0}^{1} ( {2t}^{2} - 3t + 4) .dt$

$\longrightarrow\displaystyle \sf x=\left. \Bigg( \frac{2 {t}^{3} }{3} - \frac{3 {t}^{2} }{2} + 4t\Bigg) \right| _{0}^{1}$

$\longrightarrow\displaystyle \sf x= \frac{2 }{3} - \frac{3 }{2} + 4$

$\longrightarrow\displaystyle \sf x= \frac{2 \times 2 - 3 \times 3 }{3 \times 2} + 4$

$\longrightarrow\displaystyle \sf x= \frac{4 - 9 }{6} + 4$

$\longrightarrow\displaystyle \sf x= \frac{ - 5 }{6} + 4$

$\longrightarrow\displaystyle \sf x= \frac{ - 5 + 6 \times 4 }{6}$

$\longrightarrow\displaystyle \sf x= \frac{ - 5 + 24 }{6}$

$\longrightarrow\displaystyle \sf x= \frac{ 19}{6}$

$\longrightarrow\displaystyle \underline{ \boxed{ \orange{ \bf x \approx3.17 \: m}}}$