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Solved Problem on Kinematics

A car moves along a straight road with a speed of 200 km/h. When this car passes through another car, initially at rest at a gas station, it begins to move with a constant acceleration of 4.5 m/s

a) What is the time elapsed until the car leaving the gas station reaches the speed of 200 km/h?

b) How far are they from each other when their speeds are equal?

Problem data

- speed of car
*A*:*v*_{A}= 200 km/h; - initial speed of car
*B*:*v*_{0B}= 0; - final speed of car
*B*:*v*_{B}= 200 km/h; - acceleration of car
*B*: α_{B}= 4.5 m/s^{2}.

Solution

First, let's convert the speeds of the cars given in kilometers per hour (km/h) to meters per second (m/s) used in the

\[
v_{\text{A}}=v_{\text{B}}=200\frac{\cancel{\text{km}}}{\cancel{\text{h}}}\times \frac{1000\;\text{m}}{1\;\cancel{\text{km}}}\times \frac{1\;\cancel{\text{h}}}{3600\;\text{s}}=\frac{200}{3.6}\;\frac{\text{m}}{\text{s}}=55.6\;\text{m/s}
\]

\[ \bbox[#99CCFF,10px]
{v=v_{0}+\alpha t}
\]

For car
\[
v_{\text{B}}=v_{\text{0B}}+\alpha_{\text{B}}t\\
55.6=0+4.5t\\
t=\frac{55.6}{4.5}
\]

\[ \bbox[#FFCCCC,10px]
{t=12.4\;\text{s}}
\]

b) The car

\[ \bbox[#99CCFF,10px]
{S=S_{0}+vt}
\]

using this expression, the data from car
\[
S_{\text{A}}=S_{\text{0A}}+v_{\text{A}}t\\
S_{\text{A}}=0+55.6 \times 12.4\\
S_{\text{A}}=689.4\;\text{m}
\]

Car
\[ \bbox[#99CCFF,10px]
{S=S_{0}+v_{0}t+\alpha \frac{t^{2}}{2}}
\]

using this expression, the data from car
\[
S_{\text{B}}=S_{\text{0 B}}+v_{\text{0 B}}t+\alpha_{\text{B}}\frac{t^{2}}{2}\\
S_{\text{B}}=0+0t+4.5 \times \frac{12.4^{2}}{2}\\
S_{\text{B}}=346\;\text{m}
\]

Thus the distance between the cars will be given by
\[
\Delta S=|S_{\text{A}}-S_{\text{B}}|\\
\Delta S=|689.4-346|
\]

\[ \bbox[#FFCCCC,10px]
{\Delta S=343.4\;\text{m}}
\]

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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .