Answer:74.33 feet per sec
Explanation:
Given
Police car is 50 feet side of road
Red car is 140 feet up the road
Distance between them is decreasing at the rate of 70 feet per sec
From figure
[tex]x^2+y^2=z^2[/tex]
[tex]z=\sqrt{140^2+50^2}[/tex]
z=148.66 feet
as the red car is moving
therefore its velocity magnitude is given by
[tex]\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]
[tex]2x\times \frac{\mathrm{d} x}{\mathrm{d} t}+0=2z\frac{\mathrm{d} z}{\mathrm{d} t}[/tex]
[tex]x\frac{\mathrm{d} x}{\mathrm{d} t}=z\frac{\mathrm{d} z}{\mathrm{d} t}[/tex]
[tex]140\times \frac{\mathrm{d} x}{\mathrm{d} t}=148.66\times 70[/tex]
[tex]\frac{\mathrm{d} x}{\mathrm{d} t}=74.33 feet\ per\ sec[/tex]
The actual speed of the red car along the road is 74.35 ft/s.
The resultant displacement of the car and the radar is calculated as follows;
[tex]c^2 = a^2 + b^2\\\\c = \sqrt{a^2 + b^2} \\\\c = \sqrt{50^2 + 140^}} \\\\c = 148.7 \ ft[/tex]
The change in the displacement of the vehicle is calculated as follows;
[tex]2c\frac{dc}{dt} = 2a\frac{da}{dt} + 2b\frac{db}{dt} \\\\c\frac{dc}{dt} = a\frac{da}{dt} + b\frac{db}{dt} \\\\148.7 \times 70 = 140\frac{da}{dt} \\\\\frac{da}{dt} = 74.35 \ ft/s[/tex]
Thus, the actual speed of the red car along the road is 74.35 ft/s.
Learn more about displacement here: https://brainly.com/question/2109763
