Respuesta :
Let s be the speed of the slowest train and t the time it takes to travel 300 miles.
We know that:
[tex]s\cdot t=300\text{ miles}[/tex]Since the second train has a speed of s+20 mph and travels the same 300 miles using 4 hours less time, then:
[tex](s+20mph)\cdot(t-4h)=300\text{ miles}[/tex]Isolate t from the first equation:
[tex]t=\frac{300\text{ miles}}{s}[/tex]Substitute the expression for t in the second equation to find an expression only in terms of s:
[tex](s+20mph)(\frac{300\text{miles}}{s}-4h)=300\text{miles}[/tex]Use the distributive property to rewrite the product of the quantities on the left hand side of the equation:
[tex]\begin{gathered} (s+20mph)(\frac{300\text{miles}}{s}-4h)=300\text{miles} \\ \Rightarrow \\ (s+20\text{mph)}\cdot\frac{300\text{miles}}{s}-(s+20\text{mph)}\cdot4h=300\text{miles} \\ \Rightarrow \\ s\cdot\frac{300\text{miles}}{s}+20\text{mph}\cdot\frac{300\text{miles}}{s}-4h\cdot s-4h\cdot20\text{mph}=300\text{miles} \end{gathered}[/tex]Simplify the products when possible. 4h times 20 mph is equal to 80 miles:
[tex]300\text{miles}+\frac{(20\text{mph)}(300\text{miles)}}{s}-4h\cdot s-80\text{miles}=300\text{miles}[/tex]Substract 300 miles from both sides of the equation:
[tex]\frac{(20\text{mph)}(300\text{miles)}}{s}-4h\cdot s-80\text{miles}=0[/tex]Multiply both sides by s:
[tex](20\text{mph)}(300\text{miles)}-4h\cdot s^2-80\text{ miles}\cdot s=0[/tex]This is a quadratic equation for s. Write the equation in standard form:
[tex]-4h\cdot s^2-80\text{ miles}\cdot s+(20\text{ mph})(300\text{miles)}=0[/tex]Use the quadratic formula to isolate s:
[tex]s=\frac{80\text{ miles}\pm\sqrt[]{(80\text{ miles})^2-4(-4h)(20mph)(300miles)}}{2(-4h)}[/tex]Observe that the term -4(-4h)(20mph)(300miles) is equal to +96000 squared miles, and 80 miles squared is equal to 6400 squared miles:
[tex]s=\frac{80\text{ miles}\pm\sqrt[]{6400+96000}\text{ miles}}{-8h}[/tex]Factoring out the units, we get:
[tex]s=\frac{80\pm\sqrt[]{102400}}{-8}\text{ mph}[/tex]Since the square root of 102400 is equal to 320:
[tex]s=\frac{80\pm320}{-8}\text{mph}[/tex]Taking the positive value of the plus/minus sign, we get:
[tex]s=\frac{80+320}{-8}\text{ mph}=\frac{400}{-8}\text{ mph }=-50\text{ mph}[/tex]Taking the negative value of the plus/minus sign, we get:
[tex]s=\frac{80-320}{-8}\text{ mph}=\frac{-240}{-8}\text{ mph}=30\text{ mph}[/tex]Since we first stated that s*t=300 miles and the time cannot be a negative number, the only acceptable answer is s=30 mph.
Since the velocity of the second train is s+20 mph and s=30mph, then the velocity of the second train is 50 mph.
Check the answer by verifying if all the conditions are satisfied. The problem says that the second train arrives 4 hours earlier.
The first train takes a time of 300 miles / 30 mph = 10 h.
The second train takes a time of 300 miles / 50 mph = 6h, 4 hours earlier.
Therefore, the velocities of the trains are 30 mph and 50 mph.
