An object's two dimensional velocity is given by $v (t) = (sin (\frac{π}{3} t), 2 cos (\frac{π}{2} t) - 3 t)$ $v(t) = ( sin(pi/3t) , 2cos(pi/2t )- 3t )$ . What is the object's rate and direction of acceleration at $t = 1$ $t=1$ ?

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An object's two dimensional velocity is given by $v (t) = (sin (\frac{π}{3} t), 2 cos (\frac{π}{2} t) - 3 t)$ $v(t) = ( sin(pi/3t) , 2cos(pi/2t )- 3t )$ . What is the object's rate and direction of acceleration at $t = 1$ $t=1$ ?

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Answer 1 · 2017-09-17T17:40:33

$a = \sqrt{\frac{π^{2}}{36} + 16}$ $a=sqrt(pi^2/36+16)$

Explanation:

$v_{x} (t) = sin (\frac{π}{3} t)$ $v_x(t)=sin(pi/3t)$ and $v_{y} (t) = 2 cos (\frac{π}{2} t) - 3 t$ $v_y(t)=2cos(pi/2t)-3t$
acceleration in X-direction
$a_{x} (t) = \frac{d}{d t} (v_{x} (t)) = \frac{d}{d t} (sin (\frac{π}{3} t)) = \frac{π}{3} cos (\frac{π}{3} t)$ $a_x(t)=d/dt(v_x(t))=d/dt(sin(pi/3t))=pi/3cos(pi/3t)$

acceleration in Y-direction
$a_{y} (t) = \frac{d}{d t} (v_{y} (t)) = \frac{d}{d t} (2 cos (\frac{π}{2} t) - 3 t)$ $a_y(t)=d/dt(v_y(t))=d/dt(2cos(pi/2t)-3t)$
$a_{y} (t) = - 2 (\frac{π}{2}) sin (\frac{π}{2} t) - 3 = - π sin (\frac{π}{2} t) - 3$ $a_y(t)=-2(pi/2)sin(pi/2t)-3=-pisin(pi/2 t)-3$

at t=1 sec
$a_{x} (1) = \frac{π}{3} cos (\frac{π}{3} (1)) = \frac{π}{3} cos (\frac{π}{3}) = \frac{π}{3} (\frac{1}{2}) = \frac{π}{6}$ $a_x(1)=pi/3cos(pi/3(1))=pi/3 cos(pi/3)=pi/3(1/2)=pi/6$
$a_{y} (1) = - π sin (\frac{π}{2} (1)) - 3 = - π sin (\frac{π}{2}) - 3 = - π - 3$ $a_y(1)=-pisin(pi/2 (1))-3=-pisin(pi/2)-3=-pi-3$
$a_{y} (1) = - π - 3$ $a_y(1)=-pi-3$

Hence Acceleration $a = \sqrt{{(a_{x} (t))}_{x}^{2} + {(a_{y} (t))}_{y}^{2}}$ $a=sqrt((a_x(t))^2+(a_y(t))^2$
$a = \sqrt{{(\frac{π}{6})}^{2} + {(- π - 3)}^{2}} = \sqrt{\frac{π^{2}}{36} + π^{2} + 6 π + 9} \approx 6.1638$ $a=sqrt((pi/6)^2+(-pi-3)^2)=sqrt(pi^2/36+pi^2+6pi+9)approx6.1638$

direction is ${tan}^{- 1} (\frac{a_{y}}{a_{x}}) = {tan}^{- 1} (\frac{- π - 3}{\frac{π}{6}}) \approx - {85.13}^{o}$ $tan^-1(a_y/a_x)=tan^-1((-pi-3)/(pi/6))approx-85.13^o$ from Positive X- axis

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An object's two dimensional velocity is given by $v (t) = (sin (\frac{π}{3} t), 2 cos (\frac{π}{2} t) - 3 t)$ $v(t) = ( sin(pi/3t) , 2cos(pi/2t )- 3t )$ . What is the object's rate and direction of acceleration at $t = 1$ $t=1$ ?

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An object's two dimensional velocity is given by v(t)=(sin(π3t),2cos(π2t)−3t)v(t) = ( sin(pi/3t) , 2cos(pi/2t )- 3t ). What is the object's rate and direction of acceleration at t=1t=1 ?

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An object's two dimensional velocity is given by $v (t) = (sin (\frac{π}{3} t), 2 cos (\frac{π}{2} t) - 3 t)$ $v(t) = ( sin(pi/3t) , 2cos(pi/2t )- 3t )$ . What is the object's rate and direction of acceleration at $t = 1$ $t=1$ ?