Answer:
The rate at which the distance between the cars increasing two hours later=52mi/h
Step-by-step explanation:
Let
Speed of one car, x'=48 mi/h
Speed of other car, y'=20 mi/h
We have to find the rate at which the distance between the cars increasing two hours later.
After 2 hours,
Distance traveled by one car
[tex]x=48\times 2=96 mi[/tex]
Using the formula
[tex]Distance=Time\times speed[/tex]
Distance traveled by other car
[tex]y=20\times 2=40 mi[/tex]
Let z be the distance between two cars after 2 hours later
[tex]z=\sqrt{x^2+y^2}[/tex]
Substitute the values
[tex]z=\sqrt{(96)^2+(40)^2}[/tex]
z=104 mi
Now,
[tex]z^2=x^2+y^2[/tex]
Differentiate w.r.t t
[tex]2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}[/tex]
[tex]z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}[/tex]
Substitute the values
[tex]104\frac{dz}{dt}=96\times 48+40\times 20[/tex]
[tex]\frac{dz}{dt}=\frac{96\times 48+40\times 20}{104}[/tex]
[tex]\frac{dz}{dt}=52mi/h[/tex]
Hence, the rate at which the distance between the cars increasing two hours later=52mi/h
