The equation of motion of a particle started at t = 0 is given by x = 5 sin (20t + π/3), where x is in centimetre and t in second. When does the particle
(a) first come to rest
(b) first have zero acceleration
(c) first have maximum speed?

Question

The equation of motion of a particle started t-0 is given by displaystyle x-5sin left - 20t-frac-pi -3- right - where x is in cm and t in sec- When does the particle first have maximum speed-

Toppr Admin · Accepted Answer

X-5sin-20t-x3C0-3-v-dxdt-100cos-20t-x3C0-3-a-dvdt-x2212-2000sin-20t-x3C0-3-v-0 for the first time when-20t-x3C0-3-x3C0-2-x2234-xA0-xA0- t-x3C0-120 s

The equation of motion of a particle started at $t = 0$ is given by $x = 5 sin (20 t + \frac{π}{3})$ , where $x$ is in cm and $t$ in sec. When does the particle first have maximum speed?

$X = 5 sin (20 t + \frac{π}{3})$
$v = \frac{d x}{d t} = 100 cos (20 t + \frac{π}{3})$
$a = \frac{d v}{d t} = - 2000 sin (20 t + \frac{π}{3})$
v=0 for the first time when,
$20 t + \frac{π}{3} = \frac{π}{2}$
$∴$ $t = \frac{π}{120}$ s

The equation of motion of a particle started at t=0 is given by x=5sin(20t+π3), where x is in cm and t in sec. When does the particle first have maximum speed?

X=5sin(20t+π3)v=dxdt=100cos(20t+π3)a=dvdt=−2000sin(20t+π3)v=0 for the first time when,20t+π3=π2∴ t=π120 s

The equation of motion of a particle started at $t = 0$ is given by $x = 5 sin (20 t + \frac{π}{3})$ , where $x$ is in cm and $t$ in sec. When does the particle first have maximum speed?

$X = 5 sin (20 t + \frac{π}{3})$
$v = \frac{d x}{d t} = 100 cos (20 t + \frac{π}{3})$
$a = \frac{d v}{d t} = - 2000 sin (20 t + \frac{π}{3})$
v=0 for the first time when,
$20 t + \frac{π}{3} = \frac{π}{2}$
$∴$ $t = \frac{π}{120}$ s