An object's two dimensional velocity is given by $v (t) = (\sqrt{t^{2} - 1} - 2 t, t^{2} - 5)$ $v(t) = ( sqrt(t^2-1)-2t , t^2-5)$ . What is the object's rate and direction of acceleration at $t = 2$ $t=2$ ?

Question

1804 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-06T20:15:14

$∣ ∣ \to a ∣ ∣ \approx 4.09 {ms}^{- 2}$ $|veca| ~~ 4.09"ms"^-2$
$θ \approx 1.779 rad or {101.9}^{\circ}$ $theta ~~ 1.779"rad or "101.9^@$

Here we have velocity along $x$ $x$ direction:

$v_{x} (t) = \sqrt{t^{2} - 1} - 2 t$ $v_x(t)=sqrt (t^2-1)-2t$

$:.$ acceleration along $x$ direction:

$a_x(t)=(dv_x)/(dt)=t/(sqrt (t^2-1))-2$

At $t=2s$ :

$a_x(2"s")=(2/ sqrt3-2)"ms"^-2$

Similarly

$a_y(t)=2t$

At $t=2s$ :

$a_y(2"s")=4"ms"^-2$

The acceleration vector at $t = 2$ , is $veca = 2(sqrt3/3-1)hati+ 4hatj$

The acceleration is the magnitude of this vector:

$|veca| = sqrt((2(sqrt3/3-1))^2+4^2)$

$|veca| ~~ 4.09"ms"^-2$

The direction is:

$theta = tan^-1(4/(2(sqrt3/3-1)))+ pi$

Please notice that we have added $pi$ because the vector points to the 2nd quadrant.

$theta ~~ 1.779"rad or "101.9^@$

An object's two dimensional velocity is given by v(t) = ( sqrt(t^2-1)-2t , t^2-5)v(t)=(√t2−1−2t,t2−5)v(t) = ( sqrt(t^2-1)-2t , t^2-5). What is the object's rate and direction of acceleration at t=2 t=2t=2 ?