Solved Problem on Kinematics
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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/s2 until it reaches the speed of 200 km/h. Find
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:    vA = 200 km/h;
• initial speed of car B:    v0B = 0;
• final speed of car B:    vB = 200 km/h;
• acceleration of car B:    αB = 4.5 m/s2.
Problem diagram

figura 1

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 International System of Units (S.I.)
$v_{\text{A}}=v_{\text{B}}=200\frac{\cancel{\text{km}}}{\cancel{\text{h}}}\times \frac{1000\;\text{m}}{1\;\cancel{\text{km}}}\times \frac{1\;\cancel{\text{h}}}{3600\;\text{s}}=\frac{200}{3.6}\;\frac{\text{m}}{\text{s}}=55.6\;\text{m/s}$
a) The equation of velocity as a function of time is given by

$\bbox[#99CCFF,10px] {v=v_{0}+\alpha t}$
For car B we have
$v_{\text{B}}=v_{\text{0B}}+\alpha_{\text{B}}t\\ 55.6=0+4.5t\\ t=\frac{55.6}{4.5}$
$\bbox[#FFCCCC,10px] {t=12.4\;\text{s}}$

b) The car A moves with constant speed, the equation of displacement as a function of time is given by

$\bbox[#99CCFF,10px] {S=S_{0}+vt}$
using this expression, the data from car A and the time interval found in item (a), we have
$S_{\text{A}}=S_{\text{0A}}+v_{\text{A}}t\\ S_{\text{A}}=0+55.6 \times 12.4\\ S_{\text{A}}=689.4\;\text{m}$
Car B moves with constant acceleration, the equation of displacement as a function of this time is given by
$\bbox[#99CCFF,10px] {S=S_{0}+v_{0}t+\alpha \frac{t^{2}}{2}}$
using this expression, the data from car B and the time interval found in item (a), we have
$S_{\text{B}}=S_{\text{0 B}}+v_{\text{0 B}}t+\alpha_{\text{B}}\frac{t^{2}}{2}\\ S_{\text{B}}=0+0t+4.5 \times \frac{12.4^{2}}{2}\\ S_{\text{B}}=346\;\text{m}$
Thus the distance between the cars will be given by
$\Delta S=|S_{\text{A}}-S_{\text{B}}|\\ \Delta S=|689.4-346|$
$\bbox[#FFCCCC,10px] {\Delta S=343.4\;\text{m}}$