An object's two dimensional velocity is given by $v (t) = (sin (\frac{π}{3} t), 2 cos (\frac{π}{2} t) - t)$ $v(t) = ( sin(pi/3t) , 2cos(pi/2t )- t )$ . What is the object's rate and direction of acceleration at $t = 5$ $t=5$ ?

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Answer 1 · 2018-01-04T10:04:27

The rate of acceleration is $= 4.17 m s^{- 2}$ $=4.17ms^-2$ in the direction ${97.2}^{\circ}$ $97.2^@$ anticlockwise from the $x-axis$ $"x-axis"$

The acceleration is the derivative of the velocity

$v (t) = (sin (\frac{π}{3} t), 2 cos (\frac{π}{2} t) - t)$ $v(t)=(sin(pi/3t),2cos(pi/2t)-t)$

The acceleration is

$a(t)=v'(t)=(pi/3cos(pi/3t), -pisin(pi/2t)-1)$

When $t=5$

$a(5)=v'(5)=(pi/3cos(pi/3*5), -pisin(pi/2*5)-1)$

$=(0.52, -4.14)$

The rate of acceleration is

$||a(5)||=sqrt((0.52)^2+(-4.14)^2)=4.17ms^-2$

And the direction is

$theta=arctan(-4.14/0.52)=97.2^@$ anticlockwise from the $"x-axis"$

An object's two dimensional velocity is given by v(t) = ( sin(pi/3t) , 2cos(pi/2t )- t )v(t)=(sin(π3t),2cos(π2t)−t)v(t) = ( sin(pi/3t) , 2cos(pi/2t )- t ). What is the object's rate and direction of acceleration at t=5 t=5t=5 ?