An object's two dimensional velocity is given by $v(t) = ( e^t-2t , t-4e^2 )$ . What is the object's rate and direction of acceleration at $t=1$ ?

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Answer 1 · 2017-06-29T13:07:44

The rate of acceleration is $=1.23ms^-2$ in the direction $=54.2º$ anticlockwise from the x-axis

The acceleration is the derivative of the velocity

$v(t)=(e^t-2t,t-4e^2)$

$a(t)=v'(t)=(e^t-2,1)$

So,

when $t=1$

$a(1)=(e-2,1)=(0.72,1)$

The rate of acceleration is

$=||a(1)||=sqrt(0.72^2+(1)^2)=sqrt1.52=1.23$

The direction is $arctan(1/0.72)=54.2º$ anticlockwise from the x-axis

An object's two dimensional velocity is given by v(t) = ( e^t-2t , t-4e^2 )v(t) = ( e^t-2t , t-4e^2 ). What is the object's rate and direction of acceleration at t=1 t=1 ?