Q19P - Fundamentals Of Physics

Question

At a certain time, a particle had a speed of$\mathbf{18}\mathbf{m}\mathbf{/}\mathbf{s}$ in positive x direction, and 2.4 s later, its speed was 30 m/s in the opposite direction. What is the average acceleration of particle during this 2.4 s interval?

Step-by-Step Solution

Answer

Average acceleration of particle during $2.4 \mathrm{s}$ is -$20\mathrm{m}/{\mathrm{s}}^2$  

Step 1: Given

$$\begin{gathered} {\mathrm{v}}_{\mathrm{i}}=+18\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{f}}=-3.0 \mathrm{m}/\mathrm{s} \end{gathered}$$ 

 

 ${\mathrm{t}}_1=0\mathrm{s}$, (instance at which we started observing the particle) and

${\mathrm{t}}_2=2.4 \mathrm{s}$ 

Step 2: To understanding the concept

The problem deals with the average acceleration which refers to the rate at which the velocity changes. Average acceleration can be found using the formula for the average acceleration and given data.

 

Formula:

 

The average acceleration is given by,

$a_{avg}=\frac{∆V}{∆t}$ 

Step 3: the average acceleration of particle during this 2.4 s interval

Using average acceleration equation we get,

$$\begin{gathered} {\mathrm{a}}_{\mathrm{avg}}=\frac{∆\mathrm{V}}{∆\mathrm{t}} \\ =\frac{{\mathrm{v}}_{\mathrm{f}}\left({\mathrm{t}}_2\right)-{\mathrm{v}}_{\mathrm{i}}\left({\mathrm{t}}_{\mathrm{i}}\right)}{\left(2.4-0\right)} \\ =-20 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$ 

Average acceleration calculated is $-20\mathrm{m}/{\mathrm{s}}^2$ and the magnitude is $20\mathrm{m}/{\mathrm{s}}^2$.