An object's two dimensional velocity is given by $v (t) = (2 t^{2}, - t^{3} + 4 t)$ $v(t) = ( 2t^2 , -t^3 +4t )$ . What is the object's rate and direction of acceleration at $t = 7$ $t=7$ ?

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Answer 1 · 2017-05-13T09:13:41

The rate of acceleration is $= 145.72 m s^{- 1}$ $=145.72ms^-1$ and the direction is $= 169.3$ $=169.3$ º anticlockwise from the x-axis.

The acceleration is the derivative of the velocity.

$v (t) = (2 t^{2}, - t^{3} + 4 t)$ $v(t)=(2t^2, -t^3+4t)$

$a(t)=v'(t)=(4t, -3t^2+4)$

Therefore,

$a(7)=(4*7, -3*49+4)=(28,-143)$

The rate of acceleration is

$||a(4)||=sqrt(28^2+(-143)^2)$

$=sqrt21233$

$=145.72ms^-2$

The direction is

$theta=arctan(-28/143)$

$theta$ lies in the 2nd quadrant

$theta=180-10.7=169.3º$

An object's two dimensional velocity is given by v(t) = ( 2t^2 , -t^3 +4t )v(t)=(2t2,−t3+4t)v(t) = ( 2t^2 , -t^3 +4t ). What is the object's rate and direction of acceleration at t=7 t=7t=7 ?