Question

A particle moving with velocity v=20 sinπt what is the average velocity of psrticle in 0 to 1 sec

Answer 1

Explanation:

answer x=40/π .v=dx/dt=sin(πt)

Answer 2

Given:

• Velocity of particle is v = 20 sin(πt).

To find:

• Average velocity from t = 0 to 1 sec ?

Calculation:

The average velocity can be calculated using CALCULUS as follows :

$\boxed{avg. \: v = \frac{ \displaystyle \int_{0}^{1}v \: dt}{ \displaystyle \int_{0}^{1} \: dt} }$

• Now, putting necessary values :

$\implies \:avg. \: v = \frac{ \displaystyle \int_{0}^{1}20 \sin(\pi t) \: dt}{ \displaystyle \int_{0}^{1} \: dt}$

$\implies \:avg. \: v = 20 \: \frac{ \displaystyle \int_{0}^{1} \sin(\pi t) \: dt}{ \displaystyle \int_{0}^{1} \: dt}$

$\implies \:avg. \: v = - 20 \: \dfrac{ \displaystyle \frac{ \cos(\pi t) }{\pi} \bigg|_{0}^{1}}{ 1 - 0}$

$\implies \:avg. \: v = 20 \: \dfrac{ \dfrac{ \cos(\pi t) }{\pi} \bigg|_{1}^{0}}{ 1 - 0}$

$\implies \:avg. \: v = 20 \: \dfrac{ \dfrac{ \cos(0) - \cos(\pi) }{\pi} }{ 1 - 0}$

$\implies \:avg. \: v = 20 \: \dfrac{ \dfrac{ \cos(0) - \cos(\pi) }{\pi} }{ 1 }$

$\implies \:avg. \: v = 20 \times \dfrac{ 1 - ( - 1) }{ \pi }$

$\implies \:avg. \: v = 20 \times \dfrac{ 2 }{ \pi }$

$\implies \:avg. \: v = \dfrac{ 40 }{ \pi }$

$\implies \:avg. \: v = 12.73 \: m {s}^{ - 1}$

So, average velocity is 40/π or 12.73 m/s