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 $t=2$ ?

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Answer 1 · 2017-11-05T07:07:26

The rate of acceleration is $=7.32ms^-2$ in the direction $=10.51^@$ anticlockwise from the x-axis

The acceleration is the derivative of the velocity.

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

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

Therefore,

when $t=2$

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

The rate of acceleration is

$||a(t)||=sqrt((e^2-2)^2+(1)^2)=7.32ms^-2$

The direction is

$theta=arctan(1/(e^2-2))=arctan(0.18)=10.51^@$ 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=2 t=2 ?