Q35P - Fundamentals Of Physics

Question

Figure shows a red car and a green car that move toward each other. The other figure is a graph of their motion, showing their positions ${\mathbf{x}}_{\mathbf{g}\mathbf{0}}\mathbf{=}\mathbf{270}\mathbf{m}$ and  ${\mathbf{x}}_{\mathbf{r}\mathbf{0}}\mathbf{35}\mathbf{m}\mathbf{=}\mathbf{-}\mathbf{ }$ at time  $\mathbf{t}\mathbf{=}\mathbf{0}$.The green car has a constant speed of 20.0 m/s and the red car begins from rest. What is the acceleration magnitude of the red car?

Step-by-Step Solution

Answer

Acceleration of the red car is $0.90\mathrm{m}/{\mathrm{s}}^2$

Step 1: Given information

$$\begin{gathered} x_{g0}=270\hspace{0.33em}m \\ {x}_{r0}=-35.0\hspace{0.33em}m \\ {V}_g=20.0\hspace{0.33em}m/s \end{gathered}$$

Step 2: Concept and formula used in the given question

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration. Using the formula for the second kinematic equation, the equation for the position of both the red and the green car can be written.By solving these equations simultaneously, the acceleration of the red car can be found.

Formula:

The displacement in kinematic equation can be expressed as,

 $\mathrm{\Delta x}={\mathrm{v}}_0\mathrm{t}+\frac{1}{2} {\mathrm{at}}^2$

Step 3: Calculation for theacceleration magnitude of the red car

The position of the cars as function of time is given by

 ${\mathrm{x}}_{\mathrm{g}}\left(\mathrm{t}\right)={\mathrm{x}}_{\mathrm{g}0}+{\mathrm{v}}_{\mathrm{g}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{g}}{\mathrm{t}}^2$

As acceleration ${\mathrm{a}}_{\mathrm{g}}=0 \mathrm{m}/{\mathrm{s}}^2$ , position ${\mathrm{x}}_{\mathrm{g}0}=270\hspace{0.33em}\mathrm{m}$ at time $\mathrm{t}=0\hspace{0.33em}\mathrm{s}$  

 $=(270 \hspace{0.33em}\mathrm{m})-\frac{20\hspace{0.33em}\mathrm{m}}{\mathrm{s}}\times \hspace{0.33em}\mathrm{t}$

As position ${\mathrm{x}}_{\mathrm{r}0}=-35.0\hspace{0.33em}\mathrm{m},{\mathrm{v}}_{\mathrm{r}}=0 \mathrm{m}/\mathrm{s}$  at time $\mathrm{t}=0\hspace{0.33em}\mathrm{s}$

 $$\begin{gathered} {\mathrm{x}}_{\mathrm{r}}\left(\mathrm{t}\right)={\mathrm{x}}_{\mathrm{r}0}+{\mathrm{v}}_{\mathrm{r}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{r}}{\mathrm{t}}^2 \\ {\mathrm{x}}_{\mathrm{r}}\left(\mathrm{t}\right)=(-35.0\hspace{0.33em}\mathrm{m})+\frac{1}{2}\times {\mathrm{a}}_{\mathrm{r}}\times {\mathrm{t}}^2 \end{gathered}$$

 

The graph shows two cars to pass each other at . This implies 

 $(270\hspace{0.33em}\mathrm{m})-\frac{20\hspace{0.33em}\mathrm{m}}{\mathrm{s}}12.0\hspace{0.33em}\mathrm{s}=(-35.0\hspace{0.33em}\mathrm{m})+\frac{1}{2}{\mathrm{a}}_{\mathrm{r}}(12.0{)}^2$

This equation gives  ${\mathrm{a}}_{\mathrm{r}}=0.90\hspace{0.33em}\mathrm{m}/{\mathrm{s}}^2$

Therefore, the acceleration of red car, ${\mathrm{a}}_{\mathrm{r}}=0.90\hspace{0.33em}\mathrm{m}/{\mathrm{s}}^2$ .