An object's velocity is given by $v(t) = (t^2 +t +1 , t^3- 3t )$ . What is the object's rate and direction of acceleration at $t=6$ ?

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An object's velocity is given by $v(t) = (t^2 +t +1 , t^3- 3t )$ . What is the object's rate and direction of acceleration at $t=6$ ?

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Answer 1 · 2016-04-23T07:26:24

$vecv(t)=(t^2+t+1)hati+(t^3-3t)hatj$
Differentiating w.r.t. twe have acceleration
$veca(t)=d/(dt)(v(t))=(2t+1)hati+(3t^2-3)hatj$
Acceleration at t=6 is
$veca(6)=(2xx6+1)hati+(3xx6^2-3)hatj=13hati+105hatj$
Magnitude of acceleration
$|veca(6)|=sqrt(13^2+105^2)=105.8$
Direction = $tan^-1(105/13)~~83^o$ with the x-axis

Answer 2 · 2016-04-23T07:38:46

$a=105,80 " "(unit)/s^2$

Explanation:

$a(t)=d/(d t) v(t)" derivative of v(t) give us a(t)"$

$"derivative of v(t) for x direction:"$
$"..................................................."$
$a_x(t)=d/(d t)(t^2+t+1)$

$a_x(t)=2t+1$

$"fill in t=6"$

$a_x(6)=2*6+1=13$

$a_x(6)=13$

$"derivative of v(t) for y direction"$
$"..................................................."$
$a_y(t)=d/(d t)(t^3-3t)$

$a_y(t)=3t^2-3$

$"fill in t=6"$

$a_y(6)=3*6^2-3=3*36-3=108-3=105$

$"acceleration is a vector quantity so that we have to add " a_x and a_y " as vector"$

enter image source here

$a=sqrt(13^2+105^2)$

$a=sqrt(169+11025)$

$a=sqrt(11194)$

$a=105,80 " "(unit)/s^2$

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An object's velocity is given by $v(t) = (t^2 +t +1 , t^3- 3t )$ . What is the object's rate and direction of acceleration at $t=6$ ?

2 Answers

Explanation:

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An object's velocity is given by v(t) = (t^2 +t +1 , t^3- 3t )v(t) = (t^2 +t +1 , t^3- 3t ). What is the object's rate and direction of acceleration at t=6 t=6 ?

2 Answers

Explanation:

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An object's velocity is given by $v(t) = (t^2 +t +1 , t^3- 3t )$ . What is the object's rate and direction of acceleration at $t=6$ ?