Question

a rocket - powered hockey puck moves on a horizontal frictionless table. the figures show graphs of $v_x$ and $v_y$, the $x$ - and $y$ - components of the pucks velocity. the puck starts at the origin. part a what is the magnitude of the pucks acceleration at $t = 5\\ s$? express your answer with the appropriate units. view available hint(s)

Explanation:

Step1: Recall acceleration - time relationship

Acceleration is the slope of the velocity - time graph. For the $v_x - t$ graph and $v_y - t$ graph, we will find the slopes at $t = 5s$.

Step2: Find $a_x$

For the $v_x - t$ graph, two points on the line are $(t_1=0,v_{x1} = 10\ m/s)$ and $(t_2 = 10\ s,v_{x2}=- 10\ m/s)$. The slope formula is $a_x=\frac{v_{x2}-v_{x1}}{t_2 - t_1}$. Substituting the values, we get $a_x=\frac{-10 - 10}{10-0}=-2\ m/s^{2}$.

Step3: Find $a_y$

For the $v_y - t$ graph, two points on the line are $(t_1 = 0,v_{y1}=0)$ and $(t_2 = 10\ s,v_{y2}=-10\ m/s)$. Using the slope formula $a_y=\frac{v_{y2}-v_{y1}}{t_2 - t_1}$, we have $a_y=\frac{- 10-0}{10 - 0}=-1\ m/s^{2}$.

Step4: Calculate the magnitude of acceleration

The magnitude of the acceleration vector $\vec{a}$ is given by $a=\sqrt{a_x^{2}+a_y^{2}}$. Substituting $a_x=-2\ m/s^{2}$ and $a_y=-1\ m/s^{2}$, we get $a=\sqrt{(-2)^{2}+(-1)^{2}}=\sqrt{4 + 1}=\sqrt{5}\approx2.24\ m/s^{2}$.

Answer:

$2.24\ m/s^{2}$