Question

a particle moving in the xy - plane has velocity v=(2t i+(3 - t^2) j) m/s, where t is in s. part a what is the x component of the particles acceleration vector at t = 4 s? express your answer with the appropriate units. part b what is the y component of the particles acceleration vector at t = 4 s? express your answer with the appropriate units.

Explanation:

Step1: Recall acceleration - velocity relationship

Acceleration is the derivative of velocity with respect to time. Given $\vec{v}=(2t\hat{i}+(3 - t^{2})\hat{j})\text{ m/s}$, the $x$ - component of velocity is $v_x = 2t$ and the $y$ - component of velocity is $v_y=3 - t^{2}$.

Step2: Differentiate $v_x$ to find $a_x$

Differentiate $v_x = 2t$ with respect to $t$. Using the power - rule $\frac{d}{dt}(at^n)=nat^{n - 1}$, where $a = 2$ and $n = 1$. So, $a_x=\frac{dv_x}{dt}=\frac{d(2t)}{dt}=2\text{ m/s}^2$. The acceleration in the $x$ - direction is constant and does not depend on $t$.

Step3: Differentiate $v_y$ to find $a_y$

Differentiate $v_y = 3 - t^{2}$ with respect to $t$. Using the power - rule, $\frac{d}{dt}(3)=0$ and $\frac{d}{dt}(-t^{2})=-2t$. So, $a_y=\frac{dv_y}{dt}=-2t\text{ m/s}^2$.

Step4: Evaluate $a_y$ at $t = 4\text{ s}$

Substitute $t = 4\text{ s}$ into the expression for $a_y$. $a_y=-2\times4=-8\text{ m/s}^2$.

Answer:

Part A:
$a_x = 2\text{ m/s}^2$
Part B:
$a_y=-8\text{ m/s}^2$