Answer:
The rate at which the distance between the cars increasing two hours later=52mi/h
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
![x=48* 2=96 mi](https://img.qammunity.org/2022/formulas/mathematics/college/dnpbk6l9nlypfucg8soqvsqmzoznnatjor.png)
Using the formula
![Distance=Time* speed](https://img.qammunity.org/2022/formulas/mathematics/college/f0164vwmrnb5au2j2j9toyo7pq8uxfkov3.png)
Distance traveled by other car
![y=20* 2=40 mi](https://img.qammunity.org/2022/formulas/mathematics/college/ydme5zae8u81mr2cx6lel6iqd1c6gb24qt.png)
Let z be the distance between two cars after 2 hours later
![z=√(x^2+y^2)](https://img.qammunity.org/2022/formulas/mathematics/college/92uzbntis0beyg46oizn32jfk19916opq9.png)
Substitute the values
![z=√((96)^2+(40)^2)](https://img.qammunity.org/2022/formulas/mathematics/college/hxy1lftf41yo3in44thzti73eilhmodlna.png)
z=104 mi
Now,
![z^2=x^2+y^2](https://img.qammunity.org/2022/formulas/mathematics/college/r2un97ds50i8z57w9w2jzzvumwkm7zskqa.png)
Differentiate w.r.t t
![2z(dz)/(dt)=2x(dx)/(dt)+2y(dy)/(dt)](https://img.qammunity.org/2022/formulas/mathematics/college/i77t7efwamprj781wnv95hpirtrdgc38fq.png)
![z(dz)/(dt)=x(dx)/(dt)+y(dy)/(dt)](https://img.qammunity.org/2022/formulas/mathematics/college/tcy6qskeieh2njsqww79fyrtf30ak05yki.png)
Substitute the values
![104(dz)/(dt)=96* 48+40* 20](https://img.qammunity.org/2022/formulas/mathematics/college/oyrirismaw0mwl0tjk8ddj3gq6buw0wqy8.png)
![(dz)/(dt)=(96* 48+40* 20)/(104)](https://img.qammunity.org/2022/formulas/mathematics/college/65lag0bd2f6taj8fe9n898gqs0uwi6pyh6.png)
![(dz)/(dt)=52mi/h](https://img.qammunity.org/2022/formulas/mathematics/college/yf80xzgyasg0re8kyf573zvuty6lku4k5u.png)
Hence, the rate at which the distance between the cars increasing two hours later=52mi/h