An object's two dimensional velocity is given by $v (t) = (e^{t} - 2 t, t - 4 e^{2})$ $v(t) = ( e^t-2t , t-4e^2 )$ . What is the object's rate and direction of acceleration at $t = 5$ $t=5$ ?

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An object's two dimensional velocity is given by $v (t) = (e^{t} - 2 t, t - 4 e^{2})$ $v(t) = ( e^t-2t , t-4e^2 )$ . What is the object's rate and direction of acceleration at $t = 5$ $t=5$ ?

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Answer 1 · 2018-03-07T16:49:22

The rate of acceleration is $= 146.4 m s^{- 1}$ $=146.4ms^-1$ in the direction of ${0.4}^{\circ}$ $0.4^@$ anticlockwise from the x-axis

Explanation:

The acceleration is the derivative of the velocity

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

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

When $t=5$

$a(5)=(e^5-2, 1)$

$a(5)=(146.4,1)$

The rate of acceleration is

$||a(5)||=sqrt(146.4^2+1^2)=146.4ms^-1$

The angle is

$theta=arctan(1/146.4)=0.4^@$ anticlockwise from the x-axis

Answer 2 · 2018-03-07T16:50:17

Please see below.

Explanation:

As velocity is given by $v(t)=(e^t-2t,t-4e^2)$

accelaration is $(dv)/(dt)=d/(dt)(e^t-2t)hati+d/(dt)(t-4e^2)hatj$

or $(e^t-2)hati+hatj$

Note that $-4e^2$ is a constant hence its differential is $0$ .

and at $t=5$ , accelaration is $(e^5-2)hati+hatj$

Its magnitude is $sqrt((e^5-2)^2+1)=sqrt((148.4132-2)^2+1)$

= $146.417$

and direction is $tan^(-1)(1/(e^5-2))=tan^(-1)(1/146.4132)$

= $0.39^@$

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An object's two dimensional velocity is given by $v (t) = (e^{t} - 2 t, t - 4 e^{2})$ $v(t) = ( e^t-2t , t-4e^2 )$ . What is the object's rate and direction of acceleration at $t = 5$ $t=5$ ?

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Explanation:

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An object's two dimensional velocity is given by v(t) = ( e^t-2t , t-4e^2 )v(t)=(et−2t,t−4e2)v(t) = ( e^t-2t , t-4e^2 ). What is the object's rate and direction of acceleration at t=5 t=5t=5 ?

2 Answers

Explanation:

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An object's two dimensional velocity is given by $v (t) = (e^{t} - 2 t, t - 4 e^{2})$ $v(t) = ( e^t-2t , t-4e^2 )$ . What is the object's rate and direction of acceleration at $t = 5$ $t=5$ ?