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
Apr 23, 2016

#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

Apr 23, 2016

Answer:

#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"#

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

#a=sqrt(169+11025)#

#a=sqrt(11194)#

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