# Speed of car

The car went to a city that was 240 km away. If his speed increased by 8 km/h, it would reach the finish one hour earlier. Determine its original speed.

Result

v =  40 km/h

#### Solution:

$s=vt \ \\ \ \\ 240=vt \ \\ 240=(v+8)(t-1) \ \\ \ \\ 240=(v+8)(240/v-1) \ \\ \ \\ \ \\ 240 \cdot \ v=(v+8) \cdot \ (240-v) \ \\ v^2 +8v -1920=0 \ \\ \ \\ a=1; b=8; c=-1920 \ \\ D=b^2 - 4ac=8^2 - 4\cdot 1 \cdot (-1920)=7744 \ \\ D>0 \ \\ \ \\ v_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ -8 \pm \sqrt{ 7744 } }{ 2 } \ \\ v_{1,2}=\dfrac{ -8 \pm 88 }{ 2 } \ \\ v_{1,2}=-4 \pm 44 \ \\ v_{1}=40 \ \\ v_{2}=-48 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (v -40) (v +48)=0 \ \\ v>0 \ \\ \ \\ v=v_{ 1 }=40 \ \text{km/h}$

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