Question

a bike accelerates faster, but a car goes faster. here ( f=) bikes position minus cars position. then
( \bigcirc f^{prime}(x)<0, f^{prime prime}(x)<0 )
( \bigcirc f^{prime}(x)<0, f^{prime prime}(x)>0 )
( \bigcirc f^{prime}(x)>0, f^{prime prime}(x)<0 )
( \bigcirc f^{prime}(x)>0, f^{prime prime}(x)>0 )
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Explanation:

Step1: Analyze the first - derivative

The first - derivative \(f^{\prime}(x)\) represents the rate of change of the position of the bike minus the position of the car (i.e., the relative velocity). Since the car goes faster than the bike, the distance between the bike and the car (bike's position - car's position) is decreasing. So \(f^{\prime}(x)<0\).

Step2: Analyze the second - derivative

The second - derivative \(f^{\prime\prime}(x)\) represents the acceleration of the bike minus the acceleration of the car. Since the bike accelerates faster than the car, \(f^{\prime\prime}(x)>0\) (because \(f^{\prime\prime}(x)=a_{bike}-a_{car}\) and \(a_{bike}>a_{car}\)).

Answer:

\(f^{\prime}(x)<0,f^{\prime\prime}(x)>0\) (the second option)