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A car of mass $600{\text{ }}kg$ travels along a straight horizontal road starting from a point $A$. The resistance to motion of the car is $750{\text{ }}N$.

a) The car travels from $A$ to $B$ at constant speed in $100{\text{ }}s$. The power supplied by the car’s engine is constant and equal to $30{\text{ }}kW$. Find the distance $AB$.

b) The car’s engine is switched off at $B$ and the car’s speed decreases until the car reaches $C$ with a speed of $20{\text{ }}m{\text{ }}{s^{ - 1}}$. Find the distance $BC$.

c) The car’s engine is switched on at $C$ and the power it supplies is constant and equal to $30{\text{ }}kW$. The car takes $14{\text{ }}s$ to travel from $C$ to $D$ and reaches $D$ with a speed of $30{\text{ }}m{\text{ }}{s^{ - 1}}$. Find the distance $CD$.

پاسخ تشریحی :
نمایش پاسخ

a) $DF = 30000/v$ or

$WD$ by $DF = 30000 \times 100$

$DF = R = 750{\text{ }}\left( {v = 40} \right)$ or

$WD$ by $DF = WD$ by $R = 750 \times AB$

Distance $AB$ is $4000{\text{ }}m$

b) $ - 750 = 600{\text{ }}a{\text{ }}\left( {a =  - 1.25} \right)$

${20^2} = {40^2} + 2\left( { - 1.25} \right)BC$

Distance $BC = 480{\text{ }}m$

Alternative for (b)

${\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}600\left( {{{40}^2} - {{20}^2}{\text{ }}} \right) = 750\left( {BC} \right)$

Distance $BC = 480{\text{ }}m$

c) $WD$ by engine $ = 30000 \times 14$

Gain in $KE = {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}600\left( {{{30}^2} - {{20}^2}} \right)$

$\left[ {750 \times CD = 420{\text{ }}000 - 150000} \right]$

Distance $CD$ is $360{\text{ }}m$

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