# Solving Distance Traveled Using Formulas vs. Graphs

• Callmelucky
In summary, the conversation discusses the discrepancy in calculating the distance traveled by a body in the 5th second using two different formulas - one for an accelerating body and one for the surface of a triangle. The conversation also addresses the origin of the formula for distance traveled being equal to vt/2 and the difference between distance and displacement, as well as the distinction between speed and velocity.
Callmelucky
Homework Statement
calculate the distance traveled by a body given the graph
Relevant Equations
##d=V_0\times t + \frac{1}{2}\times at^{2}##, ##d=\frac{vt}{2}##
So basically I wonder why the distance traveled by a body in the 5th second gives different results when calculated by the formula for accelerating body(##d=V_0\times t + \frac{1}{2}\times at^{2}##) and when calculated using a graph(formula for the surface of the triangle).

Here is the graph of the problem(pic below).
When I calculate distance traveled using ##d=\frac{vt}{2}## I get ##d=\frac{\Delta v\Delta t}{2}=\frac{4\times 1}{2}=2m##, and when I do the same using ##d=V_0\times t + \frac{1}{2}\times at^{2}## I get ##d=V_0\times t+\frac{1}{2}\times at^{2} = 2 + (-2)=0## because ##a=\frac{\Delta v}{\Delta t}= -4##.

If someone can please tell me where I am wrong, thank you.

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Where does the d=vt/2 expression come from ?

BvU said:
Where does the d=vt/2 expression come from ?
my textbook, it's just marked as s instead of d

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Callmelucky said:
my textbook, it's just marked as s instead of d
There must be a typo. The area of the shaded portion (a trapezoid) is
##d = \dfrac{(v_0+v)t}{2} ##

and this formula will give consistent results with the other.

-Dan

Callmelucky
topsquark said:
There must be a typo. The area of the shaded portion (a trapezoid) is
##d = \dfrac{(v_0+v)t}{2} ##

and this formula will give consistent results with the other.

-Dan
Does that then mean that the body didn't cross any distance for that one second?
One more question, I have calculated that the average speed in the first 6 seconds is 3/2 m/s but the solution in the textbook is 10/6 m/s. I used the same formula for distance traveled by accelerating the body for the first 2 seconds which is ##d=V_0\times t + \frac{1}{2}\times at^{2} = 1\times 2+\frac{1}{2}\times 0.5\times 4= 3m##, after that, distance traveled from 2nd to 5th second is 6m(v*t=2*3=6m) and distance from 5th to 6th second is 0m. So that is 9m/6sec= 3/2 m/s but the solution in the textbook is 10/9 m/s, can you please explain how?
Thank you

Callmelucky said:
One more question, I have calculated that the average speed in the first 6 seconds is 3/2 m/s but the solution in the textbook is 10/6 m/s. I used the same formula for distance traveled by accelerating the body for the first 2 seconds which is ##d=V_0\times t + \frac{1}{2}\times at^{2} = 1\times 2+\frac{1}{2}\times 0.5\times 4= 3m##, after that, distance traveled from 2nd to 5th second is 6m(v*t=2*3=6m) and distance from 5th to 6th second is 0m. So that is 9m/6sec= 3/2 m/s but the solution in the textbook is 10/9 m/s, can you please explain how?
Thank you
The distance traveled from from the 5th to the 6th second is not zero. The displacement is zero. The distance traveled is the sum of the "go" trip (5.0-5.5 s) and the "return" trip (5.5-6.0 s.) Average speed is what the odometer records over the entire trip divided by the duration of the trip.

topsquark
To add to the above, you need to discriminate distance from displacement and speed from velocity.
In a V/T graph, areas below the time axis count as positive if you want distance travelled and average speed, but negative if you want displacement and average velocity.

SammyS and topsquark

## What are the basic formulas used to calculate distance traveled?

The basic formula to calculate distance traveled is $$\text{Distance} = \text{Speed} \times \text{Time}$$. For objects in uniform motion, this straightforward equation helps determine the distance covered. For objects with varying speeds, integrating the speed over time or using the kinematic equations may be necessary.

## How can a distance-time graph be used to determine distance traveled?

A distance-time graph plots distance on the y-axis and time on the x-axis. The slope of the line on this graph represents the speed. By looking at the graph, one can determine the distance traveled over a specific time period by reading the corresponding y-values. The area under a speed-time graph also represents the distance traveled.

## What are the advantages of using formulas over graphs for calculating distance?

Using formulas to calculate distance is often quicker and more precise, especially for straightforward problems involving constant speed. Formulas are also more convenient for complex calculations involving variable speeds, where integration or differential equations might be needed.

## When is it more appropriate to use graphs instead of formulas for calculating distance?

Graphs are particularly useful when visualizing motion or when dealing with variable speeds. They provide an intuitive understanding of how distance changes over time and can be helpful in identifying patterns, trends, or anomalies that might not be immediately obvious through formulas alone.

## Can both methods be used together to solve distance traveled problems?

Yes, both methods can complement each other. For instance, a graph can provide a visual representation of the motion, helping to identify the appropriate intervals to apply specific formulas. Conversely, formulas can be used to calculate precise values that can then be plotted on a graph for a more comprehensive analysis.

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