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

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Answer 1 · 2017-07-06T21:22:39

$a = 6.52$ $"m/s"^2$

$phi = 259^"o"$

We're asked to find the object's acceleration at time $t = 1$ $"s"$ from a given velocity equation.

To do this, we need to find the object's acceleration components as a function of $t$ , via differentiation of the velocity equations:

Finding the derivative of velocity component equations, we have

$v_x = e^t - 2t^2$

$v_y = t - te^2$

$a_x = d/(dt) [e^t - 2t^2] = e^t - 4t$

$a_y = d/(dt) [t - te^2] = 1 - e^2$

At time $t = 1$ $"s"$ , the acceleration components are thus

$a_x = e^((1)) - 4(1) = -1.28$ $"m/s"^2$

$a_y = 1 - e^2 = -6.39$ $"m/s"^2$

The magnitude (rate) of the acceleration is

$a = sqrt((-1.28color(white)(l)"m/s"^2)^2 + (-6.39color(white)(l)"m/s"^2)^2) = color(red)(6.52$ $color(red)("m/s"^2$

The direction of the acceleration is

$phi = arctan((-6.39color(white)(l)"m/s"^2)/(-1.28color(white)(l)"m/s"^2)) = 78.7^"o"$ $+ 180^"o" = color(blue)(259^"o"$

measured anticlockwise from the positive $x$ -axis.

The $180^"o"$ was added to fix the calculator angle error.

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