Basic Kinematics

From MyMCAT

Introduction

How fast does an apple speed up as it drops from a tree? what path does a cannonball take as it is fired? Kinematics deals with all these concepts in terms of acceleration, velocity, and displacement.

Being able to understand mathematically, both visually and graphically, the concepts of acceleration, velocity, and displacement are essential to the MCAT. The variety of questions that can be asked involving these concepts make it applicable to almost all passages. As such, one should be familiar with all kinematics concepts and be confident in applying this knowledge to unexpected or unknown subject matter.

Displacement

Displacement is a vector value, measuring the difference from the initial position, d₁ to the final position, d₂. It is measured in meters and should not be confused with distance. Distance measures the total path the moving object took, so if a car goes forward 10 meters and back 5 meters, the final displacement is only 5 meters, but the distance is 15 meters.

Consider the following example. A student first walks 1 km north and then 2 km east. Then the student walks 1 km south and 2 km west. Where is the student located? At the exact same spot they started. Thus, the student's displacement is zero as their initial position and final position are the same (and the vector between them is nothing). What is the distance traveled however? Well the actual path of the student was that of a rectangle, and if we plot their path, the student went 1 + 2 + 2 + 1km , or 6km, in total.

More formally, we can say that

distance is a scalar and refers to "how much ground an object has covered" during its motion.
displacement is a vector which says "how far out of place is the object and in what direction is it" during its motion.

30 km west; 150km

30 km east; 150km

150 km east; 30 km

150 km west; 30 km

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If a car travels 60km and then another 90km, regardless of the directions it has traveled a total of 60+90 or 150km. This is the distance. The displacement on the other hand is a measure of the change in location from the starting position. Traveling 60km one way and then 90km the exact opposite direction implies that from the starting point it has really only gone 30km. Which way? well it has traveled more west than east (90>60) therefore we can conclude that it has gone west 30km from the starting position.

2.3 km; 31 degrees north of east

2.3 km; 31 degrees east of north

5 km; 31 degrees north of east

5 km; 31 degrees east of north

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The boat is traveling north but the river is pushing east, thus we can consider this a two dimensional problem where we will have to add vectors. Because the two vectors make a right triangle, we can use basic trigonometry to solve for the total distance. One edge of the right triangle will be 2km, the other will be 1km, now we must solve for the hypotenuse. 2² + 1² is 5, thus the length of the longest side is the squareroot of 5, which is about 2.23. In general, we would also have to find the angle/direction of the combined vector to give a complete answer, but in this case all the answers say that it is 31 degrees. If we used our rules for trig we could verify this, but instead let us just determine what the direction should be. If we travel 2km north and 1 km east, we will be more north than west and as such the angle made off of north should be smaller than the alternative. Therefore the most likely answer would be 31 degrees east of north.

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Velocity

Velocity is the rate at which displacement changes with time. It is also a vector and it is measured in meters per second, m/s.

The average velocity can be found using the following equation:

$v = \frac{d}{t}$

Note however that this assumes there is no acceleration, or if in the case that the acceleration is constant, the average velocity. It will not however tell you the instaneous velocity because if there is acceleration then the velocity is always changing and the formula does not capture this.

Acceleration

Acceleration is the rate at which velocity changes. It is also a vector and conventionally it is measured in meters per second squared, m/s². On the MCAT acceleration is always zero or a constant (because otherwise calculus would be involved in most calculations). As a result of this, we can use the following equation:

$a = \frac{ \mbox{final velocity} - \mbox{initial velocity} }{\mbox{time}} = \frac{\Delta v}{t}$

9.8 m/s down

19.6 m/s down

39.2 m/s down

98 m/s down

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Because the ball is dropped, we can assume the initial velocity is zero. Thus we have time and acceleration- this is all we need to solve the problem and so we can simply arrange for velocity.

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$\begin{align} \mbox{final velocity} - \mbox{initial velocity} &= \mbox{final velocity} - 0 \\ &= \mbox{final velocity} \\ &= a \times t \\ &= 9.8 \times 2 \\ &= 19.6 \end{align}$

7200 mph² east

-7200 mph² west

-7200 mph² east

-3600 mph² west

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Given that the questions states the breaks were applied uniformly, we can assume that constant acceleration/deceleration is implied. First, lets determine the magnitude of the acceleration. Note that because the speeds were given in mph and the time was in seconds we should convert all the units to the same thing, in this case, hours will be easier as the answers are all in miles and hours. 10 seconds = 10/3600(seconds/hour) hours = 1/360 hours

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$\begin{align}a &= \frac{ \mbox{final velocity} - \mbox{initial velocity} }{time} \\ &= \frac{40 - 60}{1/360} \\ &= \frac{-20}{1/360} \\ &= -7200 \end{align}$

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Notice that if we keep the forward direction of east, then the final velocity is smaller than the initial and a negative will result. This is an important thing to consider as it tells us what the real direction must be. We were assuming everything was going east, but the negative of east is actually west. So we can either consider the answer as "negative acceleration (deceleration) east" or "positive acceleration to the west", it just depends on how you want to interpret a negative value.

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Putting the magnitude and direction together we have a valid acceleration vector.

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Relating Them All

Consider what happens when the acceleration of a car is 2m/s² and we are starting at displacement 0. After one second, the car now has a velocity of 2m/s. After another second (two seconds total) the car has a velocity of 4m/s, at second 3s the car is travelling at 6m/s and so on. If we graph this, it is easy to see that velocity and time are forming a straight line! When acceleration is constant (as always on the MCAT), the resulting graph of velocity is linear (ie a straight line).

What about the displacement however? At time 0 the displacement was 0. After the first second the car is going 2m/s, but it took time to speed up so it couldn't have gotten to 2m displacement yet so it must have traveled only a very short distance in that one second. At time t=10s, the car's velocity is 20m/s, so in one second the car can travel 20 meters! What is going on here? Well if one where to plot of the displacement they would see it grows like a parabola getting bigger and bigger with time.

While calculus is not part of the MCAT, the graphical expression of these concepts can easily be understood through the first and second integration of acceleration with respect to time.

The Equations

While the above equations define the concepts of acceleration, velocity, and displacement, we must put them together in order to get a complete understanding. The following equations (derived from calculus) allow you to answer all the kinematics concepts involving displacement, initial and final velocity, acceleration, and time.