Question

the position of an object is described by $x(t)=at^{2}-b$ and $y(t)=b - ct^{3}$, where $a = 3 m/s^{2}$, $b = 2 m$, $c = 1 m/s^{3}$, and $t$ is in seconds. the magnitude of the objects acceleration at $t = 0$ and at $t = 1 s$ are $a_{0}$ and $a_{1}$, respectively. which of the following claims about $a_{0}$ and $a_{1}$ is true?

(a) $a_{0}=a_{1}$

(b) $a_{0}>a_{1}=0$

(c) $0 = a_{0}

(d) $0 < a_{0}

Explanation:

Step1: Find acceleration - x - component

The acceleration in the x - direction $a_x$ is the second - derivative of $x(t)$. Given $x(t)=At^{2}-B$, where $A = 3\ m/s^{2}$ and $B = 2\ m$. First derivative $v_x(t)=\frac{d x(t)}{dt}=2At$, second derivative $a_x(t)=\frac{d v_x(t)}{dt}=2A$. Substituting $A = 3\ m/s^{2}$, we get $a_x(t)=6\ m/s^{2}$ (constant).

Step2: Find acceleration - y - component

The acceleration in the y - direction $a_y$ is the second - derivative of $y(t)$. Given $y(t)=B - Ct^{3}$, where $C = 1\ m/s^{3}$ and $B = 2\ m$. First derivative $v_y(t)=\frac{d y(t)}{dt}=-3Ct^{2}$, second derivative $a_y(t)=\frac{d v_y(t)}{dt}=-6Ct$.

Step3: Calculate acceleration magnitude at $t = 0$

The magnitude of acceleration $a=\sqrt{a_x^{2}+a_y^{2}}$. At $t = 0$, $a_{y}(0)=-6C\times0 = 0$, $a_x = 6\ m/s^{2}$, so $a_0=\sqrt{6^{2}+0^{2}}=6\ m/s^{2}$.

Step4: Calculate acceleration magnitude at $t = 1$

At $t = 1\ s$, $a_y(1)=-6C\times1=-6\ m/s^{2}$, $a_x = 6\ m/s^{2}$. Then $a_1=\sqrt{6^{2}+(-6)^{2}}=\sqrt{36 + 36}=\sqrt{72}=6\sqrt{2}\ m/s^{2}$.
Since $a_0 = 6\ m/s^{2}$ and $a_1=6\sqrt{2}\ m/s^{2}\approx8.49\ m/s^{2}$, we have $0 < a_0 < a_1$.

Answer:

D. $0 < a_0 < a_1$