Answered

(Note that most of the answers are algebraic expressions involving t) A car starts on a trip and travels at a speed of 55 mph. Two hours later, a second later starts on the same trip and travels at a speed of 65 mph. When the second car has been on the road trip for t hours, the first car has traveled 55(t + 2) miles and the second car has traveled 65t miles. At time t the distance between the first car and the second car is _____ miles The ratio of the distance the second car has traveled and the distance the first car has traveled is______

Answer :

Part A: Distance between the 1st and 2nd card at time t.

Distance traveled by 1st car: D1 = 55(t + 2)

Distance traveled by 2nd car: D2 = 65t

The distance between the cars:

[tex]\begin{gathered} D_2-D_1 \\ 65t-55\mleft(t+2\mright) \end{gathered}[/tex]

Using the distributive property to remove the parentheses:

[tex]\begin{gathered} 65t-55t-110 \\ 10t-110 \end{gathered}[/tex]

The distance is: 10t - 110.

Part B: Ratio between the distance from the second to the first car.

To find it, do:

[tex]\frac{D_2}{D_1}[/tex]

Then,

[tex]\begin{gathered} \frac{65t}{55(t+2)} \\ \end{gathered}[/tex]

Dividing numerator and denominator by 5:

[tex]\begin{gathered} \frac{\frac{65}{5}t}{\frac{55}{5}(t+2)} \\ \frac{13t}{11(t+2)} \end{gathered}[/tex]

The ratio is:

[tex]\frac{13t}{11(t+2)}[/tex]

In summary,

The distance is: 10t - 110.

And the ratio is:

[tex]\frac{13t}{11(t+2)_{}}[/tex]

Go Question: Other Questions