An object's two dimensional velocity is given by $v (t) = (t^{2} + 2 t, cos π t - 3 t)$ $v(t) = ( t^2 +2t , cospit - 3t )$ . What is the object's rate and direction of acceleration at $t = 7$ $t=7$ ?

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Answer 1 · 2017-03-25T10:47:13

The rate of acceleration is $= 16.28 m s^{- 2}$ $=16.28ms^-2$
in the direction $= 169.4$ $=169.4$ º

The velocity is

$v (t) = (t^{2} + 2 t, cos (π t) - 3 t)$ $v(t)=(t^2+2t,cos(pit)-3t)$

Acceleration is the first derivative of the velocity

$a (t) = \frac{d v (t)}{d t} = (2 t + 2, - π sin (π t) - 3)$ $a(t)=(dv(t))/dt=(2t+2,-pisin(pit)-3)$

When $t = 7$ $t=7$

$a (7) = \frac{d v (7)}{d t} = (2 \cdot 7 + 2, - π sin (7 π) - 3)$ $a(7)=(dv(7))/dt=(2*7+2,-pisin(7pi)-3)$

$= (16, - 3)$ $=(16,-3)$

The object rate of acceleration is

$= \sqrt{(16^{2}) + {(- 3)}^{2}}$ $=sqrt((16^2)+(-3)^2)$

$= 16.28 m s^{- 2}$ $=16.28ms^-2$

The direction is

$= arctan (- \frac{3}{16})$ $=arctan(-3/16)$

The angle is in the 2nd quadrant and is

$= 169.4$ $=169.4$ º

An object's two dimensional velocity is given by v(t) = ( t^2 +2t , cospit - 3t )v(t)=(t2+2t,cosπt−3t)v(t) = ( t^2 +2t , cospit - 3t ). What is the object's rate and direction of acceleration at t=7 t=7t=7 ?