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

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

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Answer 1 · 2016-06-09T12:40:00

$∣ ∣ \to a (7) ∣ ∣ = 145.7 u n i t s$ $|veca(7)|=145.7units$ , rounded to one decimal place
Angle with $x$ $x$ -axis $θ = {78.9}^{\circ}$ $theta=78.9^@$ , rounded to one decimal place

Explanation:

Given two dimensional velocity
$v (t) = (2 t^{2}, t^{3} - 4 t)$ $v(t)=(2t^2, t^3-4t)$
Let the two dimensions be $x and y$ $x and y$ . Writing explicitly
$\to v (t) = 2 t^{2} ˆ i + (t^{3} - 4 t) ˆ j$ $vecv(t)=2t^2hati+ (t^3-4t)hat j$

To find acceleration $a (t)$ $a(t)$ , differentiating with time
$vecv'(t)=veca(t)=d/dt(2t^2hati+ (t^3-4t)hat j)$
or $veca(t)=4thati+ (3t^2-4)hat j$
We need to find acceleration at $t=7$
or $veca(7)=4xx7hati+ (3xx7^2-4)hat j$
$=>veca(7)=28hati+ 143hat j$
Now $|veca(7)|=sqrt(28^2+143^2)$
$|veca(7)|=145.7units$ , rounded to one decimal place
If $theta$ is the angle acceleration vector makes with $x$ -axis
$theta=tan^-1 (143/28)$
$theta=78.9^@$ , rounded to one decimal place

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

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Explanation:

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An object's two dimensional velocity is given by v(t) = ( 2t^2 , t^3 - 4t )v(t)=(2t2,t3−4t)v(t) = ( 2t^2 , t^3 - 4t ). What is the object's rate and direction of acceleration at t=7 t=7t=7 ?

1 Answer

Explanation:

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