An object's two dimensional velocity is given by $v (t) = (5 t - t^{3}, 4 t^{2} - t)$ $v(t) = ( 5t-t^3 , 4t^2-t)$ . What is the object's rate and direction of acceleration at $t = 4$ $t=4$ ?

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Answer 1 · 2017-08-22T16:34:54

The rate of acceleration is $= 53 m s^{- 2}$ $=53ms^-2$ in the direction $= {144.2}^{\circ}$ $=144.2^@$ anticlockwise from the x-axis

The acceleration is the derivative of the velocity

$v (t) = (5 t - t^{3}, 4 t^{2} - t)$ $v(t)=(5t-t^3, 4t^2-t)$

$a(t)=v'(t)=(5-3t^2, 8t-1)$

When $t=4$

$a(4)=(-43,31)$

The rate of acceleration is

$||a(4)||=sqrt((-43)^2+(31)^2)=sqrt(2810)=53$

The direction is

$theta=arctan(-31/43)=144.2^@$ anticlockwise from the x-axis

An object's two dimensional velocity is given by v(t) = ( 5t-t^3 , 4t^2-t)v(t)=(5t−t3,4t2−t)v(t) = ( 5t-t^3 , 4t^2-t). What is the object's rate and direction of acceleration at t=4 t=4t=4 ?