An object's two dimensional velocity is given by #v(t) = ( t-2 , 5t^2-3t)#. What is the object's rate and direction of acceleration at #t=1 #?

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Answer 1 · 2017-02-18T09:51:53

#"acceleration"=dotv(1))=(1,7)ms^(-2)#

#" direction tan^(-1)(7)#

if velocity is given a s a function of time the acceleration is found by differentiating the velocity function.

thus.

#v(t)=(t-2,5t^2-3t)#

#a=(dv(t))/(dt)=dotv(t)=(1,10t-3)#

the acceleration is then (assuming the measurements all are SI

#dotv(1)==(1,10-3)=(1,7)ms^(-2)#

direction is #tan^(-1)(y/x)#

#" direction tan^(-1)(7)#