Answer:
30 mi/h
Step-by-step explanation:
[tex]\frac{da}{dt}[/tex] = Velocity of car A = 24 mi/h
a = Distance car A travels in 3 hours = 24×3 = 72 mi
[tex]\frac{db}{dt}[/tex] = Velocity of car B = 18 mi/h
b = Distance car B travels in 3 hours = 18×3 = 54 mi
c = Distance between A and B after 3 hours = √(a²+b²) = √(72²+54²) = 90 mi
From pythogoras theorem
a²+b² = c²
Now, differenciating with respect to time
[tex]2a\frac{da}{dt}+2b\frac{db}{dt}=2c\frac{dc}{dt}\\\Rightarrow a\frac{da}{dt}+b\frac{db}{dt}=c\frac{dc}{dt}\\\Rightarrow \frac{dc}{dt}=\frac{a\frac{da}{dt}+b\frac{db}{dt}}{c}\\\Rightarrow \frac{dc}{dt}=\frac{18\times 54+24\times 72}{90}\\\Rightarrow \frac{dc}{dt}=30\ mi/h[/tex]
∴ Rate at which distance between the cars is increasing three hours later is 30 mi/h
