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Sagot :
Given:
The distance covered by the plane, d_A=1170 miles
The distance covered by the car, d_C=260 miles
To find:
The speed of the car.
Explanation:
Let us assume that the speed of the jet and the car is S_A and S_C respectively.
Then from the question,
[tex]S_A=S_C+325\text{ }\to\text{ \lparen i\rparen}[/tex]Let us assume the time of ride in the jet and the car is t_A and t_C respectively.
Then from the question,
[tex]t_A=t_C-1\text{ }\to\text{ \lparen ii\rparen}[/tex]The speed of the jet is given by the equation,
[tex]\begin{gathered} S_A=\frac{d_A}{t_A} \\ \implies d_A=S_At_A\text{ }\to\text{ \lparen iii\rparen} \end{gathered}[/tex]On substituting the equations (i) and (ii) in equation (iii),
[tex]d_A=(S_C+325)(t_C-1)\text{ }\to\text{ \lparen iv\rparen}[/tex]The speed of the car is given by,
[tex]\begin{gathered} S_C=\frac{d_C}{t_C} \\ \implies t_C=\frac{d_C}{S_C}\text{ }\to\text{ \lparen v\rparen} \end{gathered}[/tex]On substituting the equation (v) in equation (iv),
[tex]\begin{gathered} d_A=(S_C+325)(\frac{d_{C}}{S_{C}}-1) \\ \implies d_AS_C=(S_C+325)(d_C-S_C) \end{gathered}[/tex]On substituting the known values in the above equation,
[tex]\begin{gathered} 1170S_C=(S_C+325)(260-S_C) \\ \implies1170S_C=-S_C^2-65S_C+84500 \\ \implies S_C^2+1235S_C-84500=0 \end{gathered}[/tex]On solving the above equation,
[tex]\begin{gathered} S_C=65\text{ mph or} \\ S_C=-1300\text{ mph} \end{gathered}[/tex]As the car is moving forward, S_C=65 mph
Final answer:
The speed of the car is 65 mph.
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