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

1 Answer
A08
Jun 9, 2016

#|veca(7)|=145.7units#, rounded to one decimal place
Angle with #x#-axis #theta=78.9^@#, rounded to one decimal place

Explanation:

Given two dimensional velocity
#v(t)=(2t^2, t^3-4t)#
Let the two dimensions be #x and y#. Writing explicitly
#vecv(t)=2t^2hati+ (t^3-4t)hat j#

To find acceleration #a(t)#, differentiating with time
#vecv'(t)=veca(t)=d/dt(2t^2hati+ (t^3-4t)hat j)#
or #veca(t)=4thati+ (3t^2-4)hat j#
We need to find acceleration at #t=7#
or #veca(7)=4xx7hati+ (3xx7^2-4)hat j#
#=>veca(7)=28hati+ 143hat j#
Now #|veca(7)|=sqrt(28^2+143^2)#
#|veca(7)|=145.7units#, rounded to one decimal place
If #theta# is the angle acceleration vector makes with #x#-axis
#theta=tan^-1 (143/28)#
#theta=78.9^@#, rounded to one decimal place

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