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

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Answer 1 · 2017-10-16T05:22:02

The rate of acceleration is $=1078.1ms^-2$ in the direction of $7.62^@$ anticlockwise from the x-axis

Acceleration is the derivative of the velocity

The velocity is

$v(t)=(e^t-2t^2,t^3-4t)$

The acceleration is

$a(t)=v'(t)=(e^t-4t,3t^2-4)$

Therefore, when $t=7$

$a(7)=(e^7-28,3*49-4)=(1068.6,143)$

The rate of acceleration is

$||a(7)||=sqrt(1068.6^2+143^2)=1078.1ms^-2$

The direction is

$theta=arctan(143/1068.6)=7.62^@$ anticlockwise from the x-axis

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