Answer:
[tex]\sf a. \quad \dfrac{21}{10+x}+\dfrac{21}{10-x}=5[/tex]
Step-by-step explanation:
Formula for calculating speed:
[tex]\boxed{\sf Speed=\dfrac{Distance}{Time}}[/tex]
Therefore:
[tex]\boxed{\sf Time=\dfrac{Distance}{Speed}}[/tex]
Given values:
- x = the current (in mph)
- Speed = 10 mph
- Distance = 21 miles
- Total time = 5 hours
If the boat is traveling downstream, the current (x mph) will be pushing the boat faster, and so the boat's speed will increase by x mph.
Therefore, the expression for the time it takes for the boat to travel downstream is:
[tex]\sf Time\:(downstream)=\dfrac{21}{10+x}[/tex]
If the boat is traveling upstream, the current (x mph) will be pushing against the boat and so the boat's speed will decrease by x mph.
Therefore, the expression for the time it takes for the boat to travel upstream is:
[tex]\sf Time\:(upstream)=\dfrac{21}{10-x}[/tex]
Therefore, if the total time it takes to travel downstream and upstream is 5 hours:
[tex]\implies \sf \dfrac{21}{10+x}+\dfrac{21}{10-x}=5[/tex]
To solve for x:
[tex]\implies \sf \dfrac{21}{10+x}+\dfrac{21}{10-x}=5[/tex]
[tex]\implies \sf \dfrac{21(10-x)}{(10+x)(10-x)}+\dfrac{21(10+x)}{(10-x)(10+x)}=5[/tex]
[tex]\implies \sf \dfrac{21(10-x)+21(10+x)}{(10+x)(10-x)}=5[/tex]
[tex]\implies \sf21(10-x)+21(10+x)=5(10+x)(10-x)[/tex]
[tex]\implies \sf 210-21x+210+21x=5(100-10x+10x-x^2)[/tex]
[tex]\implies \sf 420=500-5x^2[/tex]
[tex]\implies \sf 5x^2=80[/tex]
[tex]\implies \sf x^2=16[/tex]
[tex]\implies \sf x=\pm4[/tex]
Therefore, the rate of the current is 4 mph.
