Uniformly accelerated motion - Definition and examples
This article will be about The uniformly accelerated motion, the main caracteristics of this motion, with examples and formulas.
Definition of an uniformly accelerated motion
The uniformly accelerated motion is a motion that is characterized for having a movement in a straight line and a constant acceleration and different of zero, therefore the velocity of an object in this movement is constantly changing depending on the direction of the velocity.
Main characteristics of the uniformly accelerated motion: to identify when we are talking about a uniformly accelerated motion is necessary to know the parameters of this motion: Its trajectory is in a straight line which means that this is a motion that only moves in one dimension. Its acceleration is different than zero, and this is the main difference between this motion and the constant velocity motion. Its velocity is changing constantly, because of the acceleration, the velocity increases or decreases permanently, this is why in the problems of this motion there is always going to be an initial velocity and a final velocity.
The acceleration is the constant change of the velocity. The sign or the direction that a velocity or an acceleration has could give a preview of what the movement is going to be like, if both acceleration and velocity have the same sign, then the object will increase the velocity every second, but if this magnitudes have different signs (for example a positive velocity and a negative acceleration) then the object will start to slow down every second until it is at rest.
Positive acceleration
Negative acceleration
Behavior of a uniformly accelerated motion
In a reference system we usually take up and right directions as positives and down and left directions as negatives, so if an object starts at rest, and then we apply a positive acceleration to it, the object will start to advance to the right, but if we apply a negative acceleration, then the object will start to advance to the left.
When an object in uniformly accelerated motion has an acceleration in the opposite direction than the velocity, then the velocity of the object will start to decrease until it stays in rest, after that moment the object will start to advance towards the opposite direction, an example of that is when we throw something up, we apply a velocity to the object but because of the gravity it starts to decrease its velocity and for a moment it is at rest, but then it comes down to earth again, lets remember that gravity is an acceleration that affects every object on earth.
Formulas of the Uniformly accelerated motion
There are 5 formulas to solve any problem of uniformly accelerated motion, from which we can use any of those to find a variable, but we have to know which one to use, because there are some problems that gives just certain data, but the problem does not give any data that we can use in an equation, that is why we have to pick very carefully which equation we are going to use.
Vo = initial velocity, Vf = final velocity, a = acceleration, d = distance
- vf = vo + a * t
- d = vo * t + 1/2a * t2
- vf2= vo2 + 2a * d
- vo2= vf2 - 2a * d
- d = vo + vf2* t
For instance, if in a problem we have to find the final velocity of an object and we have the initial velocity, acceleration and distance, the first equation would not work for this, because we would need the time for that, the formula that we would have to use is the third formula because it has the exact same variables that we have.
Uniformly accelerated motion graphics
Position graphic
Is the positive part of a parabola.
Velocity graphic
Is a lineal graphic.
Acceleration praphic
Is a constant graphic.
Examples of the uniformly accelerated motion
Given data Data to find
Example 1: A car that is at rest starts to accelerate 2m/s, if it keeps accelerating for 10 seconds, define the velocity that the car will reach.
Data
Vo = 0 a = 2 m/s t = 10 s Vf = ?
- Fisrt we write the equation to use
- vf = vo + a * t
- In this example we don´t have to clear the equation
- vf = 0 + 2 * 10 s
- So we directly solve it
- vf = 20
Example 2: A person that starts walking at a speed of 1m/s start accelerating and 30 seconds later this person is already running at a speed of 5m/s, ¿what was the distance that this person run?.
Data
Vo = 1 m/s vf = 5 m/s t = 30 s d = ?
- We write the formula
- d = vo + vf2* t
- Then replace the data and solve it
- d = 1 + 52* 30
- d = 62* 30
- d = 3 * 30
- d = 90
Example 3: A car that is going at a speed of 33.2 m/s starts slowing down and it ends up at rest, if we know that the acceleration applied was of -2m/s ¿What was the distance that this car covered?
Data
Vo = 33.2 m/s Vf = 0 m/s a = -2m/s d = ?
- Write the formula
- vf2 = vo2 + 2a * d
- Then we clear the distance
- d = vf2 - vo22a
- then replace the data and solve
- d = 0 - 33.222(-2)
- d = -33.22-4
- d = -1102.24-4
- d = 275.56
Example 4: a light plane is flying at a velocity of 12m/s, if this accelerates and it reaches a velocity of 20m/s, and if we know that between this it covered a 300 meters distance, what was the time and what was the acceleration.
Data
Vo = 12 m/s Vf = 20 m/s d = 300 t = ? a = ?
- We write the fith formula
- d = vo + vf2* t
- Then we clear the time and solve
- t = 2dvo + vf
- t = 2(300)12 + 20
- t = 60032
- t = 18.75 s
- Now that we have the time we write the first formula and clear the acceleration
- vf = vo + a * t
- a = vf - vot
- Then replace the data and solve
- a = 20 - 1218.75
- a = 818.75
- a = 0.43
Example 5: an acceleration of 3m/s was applied to an object during 9 seconds, if we know that the object covered a distance of 103 meters, calculate the final velocity and the initial velocity.
Data
a = 3 m/s d = 103 m t = 9 s Vo = ? Vf = ?
- Write the formula
- d = vo * t + 1/2a * t2
- Clear the initial velocity
- vo = d - 1/2a * t2t
- And solve
- vo = 103 - 1/2(3) * (9)29
- vo = 103 - 1/2(3) * (81)9
- vo = 103 - 1/2(243)9
- vo = 103 - 121.59
- vo = -18.59
- vo = -2.06
- Write the first formula
- vf = vo + a * t
- Then replace the formula and solve
- vf = -2.06 + 3 * 9
- vf = -2.06 + 27
- vf = 24.94
Ejercicio 6: There is a roller coaster in an amusement park where the first 140 meters are in a straight line, if from the beginning when the car is at rest until it hits 140 meters there is a constant acceleration of 7.5 m/s, calculate the time that took the car to get to the 140 meters and the final velocity.
Datos
Vo = 0 m/s d = 140 m a = 7.5 m/s t = ? Vf = ?
- Write the equation
- vf2 = vo2 + 2a * d
- Replace the data and solve
- vf2 = vo2 + 2a * d
- vf2 = 0 + 2(7.5) * 140
- vf2 = 0 + 15 * 140
- vf = √ 2100
- vf = 45.82
- Now we find the time
- vf = vo + a * t
- t = vf - voa
- t = 45.82 - 07.5
- t = 6.11
Example 7: A satellite in space that is at rest, is going to change his position moving in a straight line , if it reaches a velocity of 3000m/s in only 0.9s, what was the acceleration applied and define the distance.
Data
Vf = 3000 m/s Vo = 0 t = 0.9 s d = ? a = ?
- First we are going to fing the distance with the fifth formula
- d = vo + vf2* t
- d = 0 + 30002* 0.9
- d = 1500 * 0.9
- d = 1350
- And then we are going to calculate the acceleration
- vf = vo + a * t
- a = vf - vot
- a = 30000.9
- a = 3333.33
Example 8: A boat that runs a lake goes at a 5m/s velocity, but if it accelerates 1m/s during 40 meters, calculate the time that it accelerated.
Data
d = 40 m Vo = 5 m/s a = 1 m/s t = ?
This example can be solved of many different ways because if we look in the formulas one that allows us to find the time having the distance, initial velocity and the acceleration, there is only one and is the second equation, but to clear the time in this formula is not an easy work, so the easiest way to do it is to find another data first with the other equations for example we could find the initial velocity with the first equation and then use the first or the fifth formula to find the time, but in this case we are going to solve it the hard way.
In this case what we do is to order the equation until it is a second grade equation and then we have to factor it.
- We write the second formula
- d = vo * t + 1/2a * t2
- 40 = 5t + 0.5t2
- Then we pass every variable to the same side
- 0.5t2 + 5t - 40 = 0
Now with the formula written this way we are going to factor the equation, just as a reminder when, we factor a second grade equation the value of the variable always is going to have 2 answers, one negative and one positive, so we are going to choose the positive one as the answer. There are many ways to factor an equation of this type, but this is kind of hard at some point so we are going to use the second grade equation
- a = 0.5, b = 5, c = -40
- t = -b ± √b2 - 4ac2a
- t = -5 ± √52 - 4(0.5)(-40)2(0.5)
- t = -5 ± √25 +801
- t = -5 ± √25 +80
- t = -5 ± √105
- t = -5 ± 10.25
- Solution for t1
- t = -5 + 10.25
- t = 5.25
- Solution for t2
- t = -5 - 10.25
- t = -15.25
- As there can not be a negative time, the answer we will take is t = 5.25