Respuesta :
Speed is given by:
[tex]V=\frac{change\text{ in position}}{change\text{ in time}}=\frac{displacement}{time}[/tex]Let's call X the speed of car B in mph, then the speed of car A (let's call it Y) is given by x+15 mph.
The problem says car A travels 252 miles in the same time car B travels 216 miles. Then:
[tex]\begin{gathered} X=\frac{216\text{ miles}}{t\text{ hours}}\text{ Eq.(1)} \\ Y=\frac{252\text{ miles}}{t\text{ hours}}\text{ Eq.(2)} \\ \text{Also we know that} \\ Y=X+15\text{ mph Eq.(3)} \\ As\text{ they travel the same time, let's solve for t in Eq(1) and Eq(2)} \\ t=\frac{216}{X} \\ t=\frac{252}{Y} \\ \text{Then if they travel the same time t=t, thus:} \\ \frac{216}{X}=\frac{252}{Y}\text{ Eq.(4)} \end{gathered}[/tex]Now, replace Eq (3) into Eq(4) and solve for X:
[tex]\begin{gathered} \frac{216}{X}=\frac{252}{X+15} \\ \frac{216(X+15)}{X}=252 \\ \frac{X+15}{X}=\frac{252}{216} \\ \text{Applying the properties of fractions} \\ \frac{X}{X}+\frac{15}{X}=1.17 \\ 1+\frac{15}{X}=1.17 \\ \frac{15}{X}=1.17-1 \\ \frac{15}{X}=0.17 \\ 15=0.17X \\ X=\frac{15}{0.17} \\ X=90 \end{gathered}[/tex]Then, car B travels at 90 mph, now replace this value into Eq (3) and solve for Y:
[tex]\begin{gathered} Y=90+15 \\ Y=105 \end{gathered}[/tex]Thus, the speed of car A is 105 mph.
Answer: the speed of car A is 105 mph and the speed of car B is 90 moh.
