Given,
The distance of the race, d=100 m
The extended distance of the race, s=200 m
The velocity of the red car throughout the race, u₁=10 m/s
The initial velocity of the blue car, u₂=0 m/s
The blue car gains a velocity of 2 m/s every second.The constant acceleration of the blue car is
hus t
[tex]\begin{gathered} a=\frac{2}{1} \\ =2\text{ m/s}^2 \end{gathered}[/tex]
As the red car maintains the same velocity, the speed when it reaches the finish line will be 10 m/s.The time it takes for the red car to reach the finish line when the race was 100 m
[tex](t_1)_{100}=\frac{d}{u_1}[/tex]
On substituting the known values,
[tex]\begin{gathered} (t_1)_{100}=\frac{100}{10} \\ =10\text{ s} \end{gathered}[/tex]
The time it takes for the red car to reach the finish line after the race is extended,
[tex](t_1)_{200}=\frac{s}{u_1}[/tex]
On substituting the known values,
[tex]\begin{gathered} (t_1)_{200}=\frac{200}{10} \\ =20\text{ s} \end{gathered}[/tex]
From the equation of motion,
The final velocity of the blue car, when it reaches the finish line of 100 m race is given by the equation of motion,
[tex]v^2_{100}=u^2_2+2ad[/tex]
Where v₁₀O is the final velocity of the blue car at the end of the 100 m race.n substituting the known values,₀
[tex]\begin{gathered} v^2_{^{}100}=0+2\times2\times100 \\ =400 \\ v=\sqrt[]{400} \\ =20\text{ m/s} \end{gathered}[/tex]
hus the speed of the blue car when it reaches the finish line of 100 m race is 20 m/s
he time it takes for the blue car to reach the end of the 100 m race is given by,
[tex]v_{100}=u_2+a(t_2)_{100}[/tex]
Where (t₂)₁₀₀ is the time it takes for the blue car to reach the end of the 100 m race.
On substituting the known values in the above equation,
[tex]\begin{gathered} 20=0+2(t_2)_{100} \\ \Rightarrow(t_2)_{100}=\frac{20}{2} \\ =10\text{ s} \end{gathered}[/tex]
hus both cars take 10 s to reach the end of the 100 m race. Thus they both reach the finitsh line together.
The time it takes for the blue car to reach the end of the 200 m race can be calculated using the equation,
[tex]s=u_2(t_2)_{200}+\frac{1}{2}a\lbrack(t_2)_{200}\rbrack^2[/tex]
Where (t₂)₂₀₀ is the time it takes for the blue car to reach the end of the 200 m race.
On substituting the known values,
[tex]\begin{gathered} 200=0+\frac{1}{2}\times2\times\lbrack(t_2)_{200}\rbrack^2 \\ \lbrack(t_2)_{200}\rbrack^2=200 \\ \Rightarrow(t_2)_{200}=\sqrt[]{200} \\ =14.14\text{ s} \end{gathered}[/tex]
hus while rthe ed car takes 200 s to reach the finish line of the 200 m race, the blue car takes 14.14 s.
Therefore, if the race was extended, the blue car will reach the finish line first.
