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

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

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Answer 1 · 2017-01-25T04:35:18

Magnitude is $15.8115$ $15.8115$ and direction is $(- {18.435}^{\circ})$ $(-18.435^@)$

Explanation:

As $v (t) = (t^{2} + 2 t, 2 cos π t - t)$ $v(t)=(t^2+2t,2cospit-t)$ ,

at $t = 3$ $t=3$ , $v (3) = (3^{2} + 2 \times 3, 2 cos 3 π - 3)$ $v(3)=(3^2+2xx3,2cos3pi-3)$

= $(9 + 6, 2 \times (- 1) - 3)$ $(9+6,2xx(-1)-3)$ = $(15, - 5)$ $(15,-5)$

Hence object's rate is $| v (3) | = \sqrt{15^{2} + 5^{2}}$ $|v(3)|=sqrt(15^2+5^2)$

= $\sqrt{250} = 5 \sqrt{10} = 5 \times 3.1623 = 15.8115$ $sqrt250=5sqrt10=5xx3.1623=15.8115$

and direction is given by ${tan}^{- 1} (\frac{- 5}{15}) = {tan}^{- 1} (\frac{- 1}{3}) = - {18.435}^{\circ}$ $tan^(-1)((-5)/15)=tan^(-1)((-1)/3)=-18.435^@$
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An object's two dimensional velocity is given by $v (t) = (t^{2} + 2 t, 2 cos π t - t)$ $v(t) = ( t^2 +2t , 2cospit - t )$ . What is the object's rate and direction of acceleration at $t = 3$ $t=3$ ?

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

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