Question

a particle moves according to the equation x = 12t², where x is in meters and t is in seconds. (a) find the average velocity for the time interval from 2.20 s to 2.95 s. m/s (b) find the average velocity for the time interval from 2.20 s to 2.50 s. m/s

Explanation:

Step1: Recall average - velocity formula

The average - velocity formula is $v_{avg}=\frac{\Delta x}{\Delta t}=\frac{x_2 - x_1}{t_2 - t_1}$, where $x_2$ and $x_1$ are the final and initial positions, and $t_2$ and $t_1$ are the final and initial times.

Step2: Calculate positions for part (a)

Given $x = 12t^{2}$. For $t_1 = 2.20\ s$, $x_1=12\times(2.20)^{2}=12\times4.84 = 58.08\ m$. For $t_2 = 2.95\ s$, $x_2=12\times(2.95)^{2}=12\times8.7025 = 104.43\ m$. Then $\Delta t=t_2 - t_1=2.95 - 2.20 = 0.75\ s$ and $\Delta x=x_2 - x_1=104.43 - 58.08 = 46.35\ m$. So, $v_{avg}=\frac{\Delta x}{\Delta t}=\frac{46.35}{0.75}=61.8\ m/s$.

Step3: Calculate positions for part (b)

For $t_1 = 2.20\ s$, $x_1 = 12\times(2.20)^{2}=58.08\ m$. For $t_2 = 2.50\ s$, $x_2=12\times(2.50)^{2}=12\times6.25 = 75\ m$. Then $\Delta t=t_2 - t_1=2.50 - 2.20 = 0.3\ s$ and $\Delta x=x_2 - x_1=75 - 58.08 = 16.92\ m$. So, $v_{avg}=\frac{\Delta x}{\Delta t}=\frac{16.92}{0.3}=56.4\ m/s$.

Answer:

(a) $61.8\ m/s$
(b) $56.4\ m/s$