Respuesta :
Answer:
75.36 mph
Explanation:
The distance between the other car and the intersection is,
[tex]x=x_{0}+V t \\ x=\frac{1}{2}+V t[/tex]
The distance between the police car and the intersection is,
[tex]y=y_{0}+V t[/tex]
[tex]y=\frac{1}{2}-40 t[/tex]
(Negative sign indicates that he is moving towards the intersection)
Therefore the distance between them is given by,
[tex]z^{2}=x^{2}+y^{2}(\text { Using Phythogorous theorem })[/tex]
[tex]z^{2}=\left(\frac{1}{2}+V t\right)^{2}+\left(\frac{1}{2}-40 t\right)^{2} \ldots \ldots \ldots(1)[/tex]
The rate of change is,
[tex]2 z \frac{d z}{d t}=2\left(\frac{1}{2}+V t\right) V+2\left(\frac{1}{2}-40 t\right)(-40)[/tex]
[tex]2 z \frac{d z}{d t}=V+2 V^{2} t-40+3200 t \ldots \ldots \ldots[/tex]
Now finding [tex]z[/tex] when [tex]t=0,[/tex] from (1) we have
[tex]z^{2}=\left(\frac{1}{2}+V(0)\right)^{2}+\left(\frac{1}{2}-40(0)\right)^{2}[/tex]
[tex]z^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ z=\sqrt{\frac{1}{2}} \approx 0.7071[/tex]
The officer's radar gun indicates 25 mph pointed at the other car then, [tex]\frac{d z}{d t}=25[/tex] when [tex]t=0,[/tex] from
From (2) we get
[tex]2(0.7071)(25)=V+2 V^{2}(0)-40+3200(0)[/tex]
[tex]2(0.7071)(25)=V+2 V^{2}(0)-40[/tex]
[tex]35.36=V-40[/tex]
[tex]V=35.36+40=75.36[/tex]
Hence the speed of the car is [tex]75.36 mph[/tex]
