This is a time and distance problem. Please help!

Question

Victoria and margaret left their apartment at the same time traveling in the same direction. Victoria drove at 55 mph and Margaret drove at 70 mph. How many hours was it before Margaret was 30 miles ahead of Victoria?

Answer 1

Since they left at the same time and travel in the same direction, one needs only consider the difference in speeds. Since Margaret is travelling 70 - 55 = 15 miles per hour faster than Victoria, she will be 15 miles farther ahead for each hour they travel. Thus, she will be 30 miles ahead in 30 / 15 = 2 hours.

Answer 2

Victoria and margaret left their apartment at the same time traveling in the same direction. Victoria drove at 55 mph and Margaret drove at 70 mph. How many hours was it before Margaret was 30 miles ahead of Victoria?

In word problems, first seek what data was given and what is the question:

Question: How many hours before Margaret was 30 miles ahead of Victoria = h

Victoria's speed = 55mph

Margaret's speed = 70mph

They drove in the same direction (an important thing to think about for speed/time/distance problems with two characters).

The formula for speed is rate = distance/time

A related formula for time = distance/rate

We want to find the time h. We know the speeds but we don't know the distance. Because we are asked the time when Margaret was 30 miles ahead of Victoria, we can consider Victoria's distance is d and Margaret's distance is d + 30.

With this information, we can plug in the formulas:

Victoria h = d/55

Margaret h = (d + 30)/70

Because the question is asking for the same amount of time concerning both Victoria and Margaret, the h in both equations are equal. Therefore we can use substitution and substitute one of the hs with its value from the other equation.

d/55 = (d + 30)/70 This is in the form of a proportion so we can cross multiply.

70d = 55(d + 30) Eliminate parenthesis by distributive property.

70d = 55d + 1650 Subtract 55d from both sides.

15d = 1650 Divide both sides by 15

d = 110

Now that we have found Victoria's distance, we can plug that value in to determine the time h.

h = d/55

h = 110/55

h = 2

It took 2 hours for Margaret get 30 miles ahead of Victoria.

Answer 3

This is a distance-rate-time problem where the basic equation will be D_M - D_V = 30 where D_v represents the distance that Victoria travels, D_m represents the distance that Margaret travels, and 30 is the distance between them after t hours, when Margaret is 30 miles ahead of Victoria. Using the given rates for each person, and the fact that they are both travelling the same amount of time, and replacing each "D" with the rate x time for each person, the equation will look like this:

70t - 55t = 30. Combining the like "t" terms, we have 15t = 30, so t = 2. Checking this in our equation, we find that it works, so after 2 hours, the distance between Victoria and Margaret will be 30 miles with Margaret in the lead.

Answer 4

distance = speed . time

The time it takes for them to be 30 miles apart, is the unknown quantity we are looking for - so we'll call that time x hours

Margaret's distance = 70 . x

Victoria's distance = 55 . x

We want to know when the difference in their respective distances will be 30 miles. So, we have:

Margaret's distance - Victoria's distance = 70x - 55x

30 = 15x

30/15 = x

Therefore, x = 2 hours.

Margaret will be 30 miles ahead of Victoria in 2 hours.

Answer 5

Let's consider Margaret's speed to be Sm= 70 mph, and Victoria's to be Sv = 55 mph.

The total distance traveled by both after "t" hours is given by: Dm (Distance traveled by Margaret) = 70·t and Dv (distance traveled by Victoria) = 55·t. So we take the difference of the distance traveled by the two and set it equal to 30 miles:

Dm - Dv = 30

70·t - 55·t = 30

And then we solve for t:

t·(70 - 55) = 30

t = 30 / (70 - 55)