# Distance and speed word problem. Introduction to vectors.

• Nirupt
In summary: You correctly identified that the key to solving this problem is setting up an equation for the difference between the distances traveled by the two cars, and then using the given information about the slower car's speed and the 15 minute time interval to solve for the distance traveled by the faster car. Great job! In summary, the problem involves two cars starting from the same point at the same time, with one traveling at a constant speed of 55 mph and the other at 70 mph. The goal is to find the distance traveled by the faster car when it has a 15 minute lead on the slower car. To solve this, we can set up an equation for the difference between the distances traveled by the two cars and use the given information to
Nirupt

## Homework Statement

Two cars start from the same point at the same time, with one car traveling at a constant speed of 55 mph, and the other car having a constant speed of 70 mph. How far has the faster car traveled when it has a 15 minute lead on the slower car?

Problems like this always throw me off.. I need help on the process of this problem.

## The Attempt at a Solution

Not sure how to set it up really. I think I have to use D=r/t

I also know the answer is in miles because it's distance, other than that I'm uncertain.

For motion at a constant speed:

distance = speed*time (d = vt)

You can use this to set up an expression for the distance vs. time of each car. Let car 1 be the v1 = 55 mph car and car 2 be the v2 = 70 mph car. At t = 0, d1 = d2 = 0 (they start at the same point).

Now write down the equations for distance vs. time for each car. How far has each one traveled after t =15 minutes?

So would that just be the velocity (70 and 55) times the time (15)?
My answer choices are
58 mi

64 mi

44 mi

72 mi

I'm thinking 44 with that information

I am guilty of misintepreting the word problem myself. For the faster car to have a 15 minute lead over the slower one, it means that the distance by which it is *gaining* over the slower car is such that it would take the slower car 15 min to make up for that lag (if it could...which it can't, because the lag grows, of course).EDIT: for clarity: enough time has elapsed that the faster car is 15 min ahead of the slow car, at slow car speed.

So, first you have to figure how an equation for the *difference* between the distance traveled by the two cars vs. time. Then you have to set this difference equal to the distance that the slower car can travel in 15 minutes. Solve for the time it took for this distance (gain) to accumulate, and figure out how far the faster car traveled in that amount of time.

Last edited:
I get the concept now just not how to get the equation setup

Nirupt said:
I get the concept now just not how to get the equation setup

Did you write down this equation?

cepheid said:
So, first you have to figure how an equation for the *difference* between the distance traveled by the two cars vs. time. Then you have to set this difference equal to the distance that the slower car can travel in 15 minutes. Solve for the time it took for this distance (gain) to accumulate, and figure out how far the faster car traveled in that amount of time.

This paragraph above basically walks you through it step by step. For the part in bold: you know an equation for d1 vs. time, and you know an equation for d2 vs time, so finding an equation for their difference is trivial.

The only way I feel I can solve for distance is multiplying both speeds by 15 but if I subtract them I get 225. And how do I know how much the slower car can go in 15 minutes?

Nirupt said:
The only way I feel I can solve for distance is multiplying both speeds by 15 but if I subtract them I get 225.

First of all, you shouldn't be getting anything nearly as large as 225. Are you remembering to include your UNITS? Your speeds are given in miles per HOUR. Your times are given in MINUTES. You need to have consistent units.

Secondly, let me go through my hint again, with one small addition:

cepheid said:
This paragraph above basically walks you through it step by step. For the part in bold: you know an equation for d1 vs. time, and you know an equation for d2 vs time, so finding an equation for their difference: d1 - d2. is trivial.

If the distance traveled by the fast car is d1, and the distance traveled by the slow car is d2, then the difference between them (how far ahead the fast car is) is d1 - d2. You have an expression for d1 vs. time. You have an expression for d2 vs. time. Therefore, you have an expression for d1 - d2 vs. time. Come on. I can't be any more explicit without just doing the calculations for you.

Nirupt said:
And how do I know how much the slower car can go in 15 minutes?

Really? You know its speed. You know the time interval (15 minutes). We already established above that distance = speed*time.

So... I think I got it.

Since the slower car is 15 minutes behind it's 55mph * 15/60 = 13.75 miles.

Which means that the faster car is 13.75 miles ahead

so R1(t) = R2(t) + 13.75
Or 70(t) - 55(t) = 13.75
= 15t = 13.75
= .917 hours
D = r*t
.917 * 70 = 64 miles (rounded)

Nirupt said:
So... I think I got it.

Since the slower car is 15 minutes behind it's 55mph * 15/60 = 13.75 miles.

Which means that the faster car is 13.75 miles ahead

so R1(t) = R2(t) + 13.75
Or 70(t) - 55(t) = 13.75
= 15t = 13.75
= .917 hours
D = r*t
.917 * 70 = 64 miles (rounded)

Looks good to me!

## 1. What is a vector?

A vector is a mathematical quantity that represents both magnitude and direction. It is typically represented by an arrow, with the length of the arrow representing the magnitude and the direction of the arrow indicating the direction.

## 2. How do you calculate distance using vectors?

To calculate distance using vectors, you can use the Pythagorean theorem. The length of the vector represents the hypotenuse of a right triangle, and the x and y components of the vector represent the other two sides. By using the Pythagorean theorem (a^2 + b^2 = c^2), you can find the length of the vector and therefore the distance.

## 3. What is the difference between distance and displacement?

Distance is the total length traveled, while displacement is the change in position from the starting point to the ending point. Distance is a scalar quantity, while displacement is a vector quantity that takes into account both magnitude and direction.

## 4. How do you calculate speed using vectors?

To calculate speed using vectors, you can use the formula speed = distance/time. Since vectors represent distance and direction, you can use the magnitude of the vector for distance and divide by the time it took to travel that distance.

## 5. What are some real-life applications of vectors?

Vectors have many real-life applications, such as navigation systems, weather forecasting, and video game graphics. They are also used in physics to describe the motion of objects and in engineering for designing structures and machines.

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