Question

Two trains made the same 300 mile run. one train traveled 20 mph faster than the other. It arrived 4 hours earlier. find the speed of each train.

Answer 1

Let s be the speed of the slowest train and t the time it takes to travel 300 miles.

We know that:

`s\cdot t=300\text{ miles}`

Since the second train has a speed of s+20 mph and travels the same 300 miles using 4 hours less time, then:

`(s+20mph)\cdot(t-4h)=300\text{ miles}`

Isolate t from the first equation:

`t=\frac{300\text{ miles}}{s}`

Substitute the expression for t in the second equation to find an expression only in terms of s:

`(s+20mph)(\frac{300\text{miles}}{s}-4h)=300\text{miles}`

Use the distributive property to rewrite the product of the quantities on the left hand side of the equation:

`\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}`

Simplify the products when possible. 4h times 20 mph is equal to 80 miles:

`300\text{miles}+\frac{(20\text{mph)}(300\text{miles)}}{s}-4h\cdot s-80\text{miles}=300\text{miles}`

Substract 300 miles from both sides of the equation:

`\frac{(20\text{mph)}(300\text{miles)}}{s}-4h\cdot s-80\text{miles}=0`

Multiply both sides by s:

`(20\text{mph)}(300\text{miles)}-4h\cdot s^2-80\text{ miles}\cdot s=0`

This is a quadratic equation for s. Write the equation in standard form:

`-4h\cdot s^2-80\text{ miles}\cdot s+(20\text{ mph})(300\text{miles)}=0`

Use the quadratic formula to isolate s:

`s=\frac{80\text{ miles}\pm\sqrt[]{(80\text{ miles})^2-4(-4h)(20mph)(300miles)}}{2(-4h)}`

Observe that the term -4(-4h)(20mph)(300miles) is equal to +96000 squared miles, and 80 miles squared is equal to 6400 squared miles:

`s=\frac{80\text{ miles}\pm\sqrt[]{6400+96000}\text{ miles}}{-8h}`

Factoring out the units, we get:

`s=\frac{80\pm\sqrt[]{102400}}{-8}\text{ mph}`

Since the square root of 102400 is equal to 320:

`s=(80\pm320)/(-8)\text{mph}`

Taking the positive value of the plus/minus sign, we get:

`s=(80+320)/(-8)\text{ mph}=(400)/(-8)\text{ mph }=-50\text{ mph}`

Taking the negative value of the plus/minus sign, we get:

`s=(80-320)/(-8)\text{ mph}=(-240)/(-8)\text{ mph}=30\text{ mph}`

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.