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## How to strategy Position and also Velocity vs. Time MCAT Questions:

**1. What"s the Nature the the Question?**

Position or velocity matches time graphs generally present some figure with X and Y axes and connected heat segments representing either the position or velocity the some moving object.

As with any kind of figure, the an initial important thing to do is to check the axes. Identify whether the graph is showing place versus time or VELOCITY versus time. The only way to gain the question right is to know what the graph is representing (or guessing right, yet 100% odds are much better than 20%). When you have established what the graph is showing, you deserve to now begin to solve the problem.## Position vs. Time Example

*Image from Algebralab.org*So you’re presented the above graph and also now need to interpret it. Together the axes show, this is representing position versus time. V a place versus time graph, over there are 5 pieces of information you have the right to obtain:

**1. Position:**your actual position at any kind of given time. To execute this, merely look in ~ the Y-value (representing the position) at any given X-value (time). You’re unlikely to get a inquiry on this however it’s essential to understand this allude nonetheless.

**2. Displacement:**This is discovered by choosing any type of two points on the graphed line and simply looking in ~ the adjust in the Y-value. Remember that displacement is a vector:direction matters.

**3. Distance:**Distance is a scalar. Direction doesn’t matter here. So to calculate the distance in between two points, you need to calculate the separation, personal, instance displacements of the individual line segments (the readjust in the Y-value) between any type of two points, and also then include up the absolute value of every those separation, personal, instance segments. In other words, if the graph reflects a optimistic displacement of 5 meters, adhered to by a negative displacement that 5 meters, the complete distance traveled is 10 meter (whereas the total displacement would have actually been zero meters).

**4. Instantaneous Velocity:**You don’t need to recognize calculus because that the MCAT, however it is beneficial to realize that the tangent line of a position versus time graph (the slope at any certain point) provides you the instantaneous velocity.

**5. Mean Velocity:**By looking in ~ the steep between any two clues (versus the tangent at a particular point as above), you can attain the average velocity between those 2 points:

## Velocity vs. Time Exam

**Image indigenous Algebralab.org*

With a velocity versus time graph, the vital things to translate are the following: **1. Velocity:** At any given time, the velocity is found by just looking in ~ the Y-axis worth at that certain X-value (time). **2. Acceleration:** The steep of velocity versus time graphs provides you acceleration. For the MCAT you normally won"t have to differentiate in between average or instantaneous acceleration. Two points will be essential for acceleration problems:

**a.** If velocity is increasing over time (i.e. The steep is positive), acceleration is positive. This means that also if the velocity is negative, if it’s becoming less negative (i.e. Walk from to speak -25 m/s come -5 m/s), the acceleration is positive. **b.** If velocity is decreasing with time (i.e. The slope is negative), acceleration is negative. This way that also if the velocity is positive, if it’s coming to be less optimistic (i.e. Walk from say +25 m/s come +5 m/s), acceleration is negative.

**3. Displacement:** This is found by looking at the area under the curve that velocity versus time graphs. Because displacement is a VECTOR (direction matters), areas that are over the X-axis that are on the optimistic velocity side will certainly be provided positive values, while locations that are listed below the X-axis on the an adverse velocity side will be given an adverse values. (Note: The area "under the curve" is the area between the duty and the X-axis).**4. Distance:** The technique to street is similar to displacement—again you need to calculate the different locations between the velocity curve and also the X-axis. However, distance is a scalar. Thus you include the absolute values of the areas (as protest to displacement where areas beneath the X-axis were assigned negative values).

## Units-Based Approach

Here"sone more strategy that have to solidify your expertise of this topic as well as give you some insurance on check day:**Always use units in graphs to your advantage.See more: A(N) Is A Group Of Words That Includes Both A Subject And A Verb.**What do I mean by this? Well, at some point when looking in ~ the slopes or locations of the curves i’ve presented, we’re performing the following mathematical operations:

**Slopes:**Here, we’re effectively splitting Y-axis values by X-axis values, ~ all, steep = "rise over run" or m = ∆y/∆x

**Areas:**Here, we’re effectively multiplying Y-axis values by X-axis values. Knowing the systems (either place or velocity top top the Y-axis and time ~ above the X-axis) and the mathematical procedure being perform (none in the case of merely reporting the worth at a given time, division of Yby Xin the case of a slope, multiplication of Yby Xin the case of an area) will therefore tell you the units of the answer.

**Slope of position versus time graph:**This is splitting position (SI units of meters) by time (SI units of seconds). M/S gives you the units of rate or velocity, which is specifically what taking the steep of place versus time graphs is providing you.

**Slope the velocity matches time graph:**This is dividing velocity (SI devices of meters/second) by time (SI devices of seconds). M/S^2 gives you the systems of—you guessed it—acceleration. And sure enough, acquisition the steep of velocity matches time gives you acceleration.

**Area that velocity matches time graph:**This is multiplying velocity (SI systems of meters/second) through time (SI systems of seconds), offering you a final unit the meters, i m sorry is the unit of street or displacement. And also that is specifically what this operation gives you.

**Area of place versus time graph:**This multiplies place (meters) through time (seconds) definition the area here has actually units the meters*seconds. This most likely doesn’t remind you of the systems of something you’re acquainted with. And that is good because this isn’t a coherent metric and isn’t something girlfriend will ever before need to or have to calculate because that the MCAT.

## Conclusion

Ultimately, if she comfortable through the above rules and also understand these topics, you will certainly be ready to tackle any kind of position or velocity matches time graph. As with most difficult concepts in physics, if you have a consistent strategy these difficulties are really solvable also if they at first seem confuse or difficult.If there are other topics you"d prefer us to malfunction for you, or if you have actually questions around this breakdown, let us understand below. Good luck!

Further reading:

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