Kinematic Equations

The Kinetic Equations

The Kinetic Equations, also called the Equations of Motion are the following five: $v_{1} = a t + v_{0}$ $r_{1} = r_{0} + v_{0} t + \frac{a t^{2}}{2}$ $r_{1} = r_{0} + \frac{(v_{1} + v_{0}) t}{2}$ ${v_{1}}^{2} = {v_{0}}^{2} + 2 a (r_{1} + r_{0})$ $r_{1} = r_{0} + v_{1} t - \frac{a t^{2}}{2}$

The parameters of the kinetic equations

$a =$ the constant acceleration
$t =$ the elapsed time
$r_{0} =$ the initial position
$r_{1} =$ the final position
$v_{0} =$ the initial velocity
$v_{1} =$ the final velocity
( $g =$ the gravitational acceleration of planet earth at your latitude and altitude e.g. 9.81 $m / s^{2}$ )

The parameters of the kinetic equations are only six. ( $g$ is a special case of $a$ ) In any given problem you will know some of them while others are unknown. You have to calculate them using the equations above.

Solving the problems you are given case-by-case

Lookup Table for all cases
A = Accelerating
B = Breaking
U = Unknown
K = Known

Case Breaking or Accelerating Distance Traveled Constant Acceleration Final Velocity Initial Velocity Elapsed Time

1 A U U K K K

2 B U U K K K

3 A K U K K U

4 B K U K K U

5 A K K U K U

6 B K K K U U

7 A K K U U U

8 A U K U K K

9 B U K K U K

10 A K U U U K

Case 1:
You are given the assignment of finding the acceleration and/or the distance traveled when a body accelerates from standstill to the final speed $v_{1}$ in $t$ seconds.
Case 2:
You are given the assignment of finding the acceleration and/or the distance traveled when a body breaks from the initial speed $v_{0}$ to standstill in $t$ seconds.

Case 1 and Case 2 are both solved the same way. The only difference between Case 1 & 2 is that in Case 1 $v_{0} = 0$ and $v_{1} > 0$ while in Case 2 $v_{0} > 0$ and $v_{1} = 0$

Using our first kinetic equation, which was $v_{1} = a t + v_{0}$

The acceleration is $a = \frac{v_{1} - v_{0}}{t}$ You can notice that if $v_{1} = 0$ the acceleration $a < 0$ , which means that it's negative. So we're breaking instead of accelerating.
If $v_{0} = 0$ instead the acceleration $a > 0$ , which means that it's positive. So we're accelerating this time.

Using our third kinetic equation, which was $r_{1} = r_{0} + \frac{(v_{1} + v_{0}) t}{2}$ If we set $r_{0}$ to zero, then $r_{1}$ is the distance traveled. $r_{1} = \frac{(v_{1} + v_{0}) t}{2}$ $v_{0}, v_{1}$ and $t$ are all known. You just have to enter the values and do the calculation.

Case 3:
You are given the assignment of finding the acceleration when a body accelerates from standstill to the final speed $v_{1}$ traveling the distance $r_{1}$ .
Case 4:
You are given the assignment of finding the acceleration when a body breaks from the initial speed $v_{0}$ to a standstill traveling the distance $r_{1}$ .

Case 3 and Case 4 are both solved the same way. The only difference between Case 3 & 4 is that in Case 3 $v_{0} = 0$ and $v_{1} > 0$ while in Case 4 $v_{0} > 0$ and $v_{1} = 0$

You can notice that if $v_{1} = 0$ the acceleration $a < 0$ , which means that it's negative. So we're breaking instead of accelerating.
If $v_{0} = 0$ instead the acceleration $a > 0$ , which means that it's positive. So we're accelerating this time.

Using our fourth kinetic equation, which was ${v_{1}}^{2} = {v_{0}}^{2} + 2 a (r_{1} + r_{0})$ If we set $r_{0}$ to zero, then $r_{1}$ is the distance traveled. Now we can isolate $a$ . $a = \frac{{v_{1}}^{2} - {v_{0}}^{2}}{2 r_{1}}$ $v_{0}, v_{1}$ and $r_{1}$ are all known. You just have to enter the values and do the calculation.

Case 5:
You are given the assignment of finding the elapsed time and/or the final speed when a body accelerates with a given acceleration from standstill to some final speed traveling the distance $r_{1}$ .
Case 6:
You are given the assignment of finding the elapsed time and/or the initial speed when a body breaks with a given acceleration, $a$ , from some initial speed to a standstill traveling the distance $r_{1}$ .

Case 5 and Case 6 are both solved the same way. The only difference between Case 5 & 6 is that in Case 5 $v_{0} = 0$ and $v_{1} > 0$ while in Case 6 $v_{0} > 0$ and $v_{1} = 0$ .

For Case 5 we can use our second kinetic equation, which was $r_{1} = r_{0} + v_{0} t + \frac{a t^{2}}{2}$ If we set $r_{0}$ to zero, then $r_{1}$ is the distance traveled. Since we start from a standstill $v_{0} = 0$ . Now we can isolate $t$ . $t = \sqrt{\frac{2 r_{1}}{g}}$ For Case 6 we can use our fifth kinetic equation. This renders the same solution as the one above. Both $r_{1}$ and $g$ are known. You just have to enter the values and do the calculation.

Regarding respectively the initial and final speed we can use our fourth kinetic equation which was: ${v_{1}}^{2} = {v_{0}}^{2} + 2 a (r_{1} + r_{0})$
In Case 5 we have the conditions that $r_{0} = 0$ and $v_{0} = 0$ . In Case 6 we have the conditions that $r_{0} = 0$ and $v_{1} = 0$ . Both of these cases renders the same solution if we disregard negative speeds. This formula is: $If v_{0} = 0 then v_{1} = \sqrt{2 a r_{1}}$ $If v_{1} = 0 then v_{0} = \sqrt{2 a r_{1}}$ Case 7:
You are given the assignment of finding the elapsed time and/or the initial speed when a body is thrown up in the air to the height of $r_{1}$ and then lands on the ground.
To find the initial speed in Case 7 we measure only from the bottom to the top of the trajectory where the body turns and goes down again. So $r_{0} = 0$ and $v_{1} = 0$ . This yields the same solutions for the initial speed as in Case 5 and 6 $v_{0} = - v_{1} = \sqrt{2 g r_{1}}$ Regarding the time, the body will travel both up and down. So the time it will be in the air will be twice that of Case 5 and 6. So for Case 7 $t = 2 \sqrt{\frac{2 r_{1}}{g}}$ Case 8:
You are given the assignment of finding the distance traveled and/or the final speed or the initial speed when a body accelerates from standstill to some final speed in a given period of $t$ seconds.
Case 9:
You are given the assignment of finding the distance traveled and/or the final speed or the initial speed when a body breaks from some initial speed to a standstill in a given period of $t$ seconds.

Case 8 and Case 9 are both solved the same way. The only difference between Case 8 & 9 is that in Case 8 $v_{0} = 0$ and $v_{1} > 0$ while in Case 9 $v_{0} > 0$ and $v_{1} = 0$

Using our first kinetic equation, which was $v_{1} = a t + v_{0}$ We can clearly see that $If v_{0} = 0 then v_{1} = a t$ $If v_{1} = 0 then v_{0} = a t$ (disregarding the signs)

For Case 8, in order to find the distance traveled we can use our second kinetic equation, which was $r_{1} = r_{0} + v_{0} t + \frac{a t^{2}}{2}$ In Case 8 we have the following conditions $r_{0} = 0$ and $v_{0} = 0$ . This yields that the distance traveled $r_{1} = \frac{a t^{2}}{2}$ Both $a$ and $t$ are known. You just have to enter the values and do the calculation.

For Case 9, in order to find the distance traveled we can use our fifth kinetic equation, which was $r_{1} = r_{0} + v_{1} t - \frac{a t^{2}}{2}$ In Case 9 we have the following conditions $r_{0} = 0$ and $v_{1} = 0$ . This yields that the distance traveled $r_{1} = \frac{a t^{2}}{2}$ (disregarding the signs)
Both $a$ and $t$ are known. You just have to enter the values and do the calculation.

Case 10:
You are given the assignment of finding the acceleration when a body accelerates from standstill to some final speed in a given period of $t$ seconds over a distance of $r_{1}$ .

For Case 10, in order to find the distance traveled we can use our second kinetic equation, which was $r_{1} = r_{0} + v_{0} t + \frac{a t^{2}}{2}$ In Case 10 we have the following conditions $r_{0} = 0$ and $v_{0} = 0$ . This yields that the distance traveled $r_{1} = \frac{a t^{2}}{2}$
Now isolate $a$ $a = \frac{2 r_{1}}{t^{2}}$ Both $r_{1}$ and $t$ are known. You just have to enter the values and do the calculation.

Case	Breaking or Accelerating	Distance Traveled	Constant Acceleration	Final Velocity	Initial Velocity	Elapsed Time
1	A	U	U	K	K	K
2	B	U	U	K	K	K
3	A	K	U	K	K	U
4	B	K	U	K	K	U
5	A	K	K	U	K	U
6	B	K	K	K	U	U
7	A	K	K	U	U	U
8	A	U	K	U	K	K
9	B	U	K	K	U	K
10	A	K	U	U	U	K

Case	Breaking or Accelerating	Distance Traveled	Constant Acceleration	Final Velocity	Initial Velocity	Elapsed Time
1	A	U	U	K	K	K
2	B	U	U	K	K	K
3	A	K	U	K	K	U
4	B	K	U	K	K	U
5	A	K	K	U	K	U
6	B	K	K	K	U	U
7	A	K	K	U	U	U
8	A	U	K	U	K	K
9	B	U	K	K	U	K
10	A	K	U	U	U	K

Case	Breaking or Accelerating	Distance Traveled	Constant Acceleration	Final Velocity	Initial Velocity	Elapsed Time
1	A	U	U	K	K	K
2	B	U	U	K	K	K
3	A	K	U	K	K	U
4	B	K	U	K	K	U
5	A	K	K	U	K	U
6	B	K	K	K	U	U
7	A	K	K	U	U	U
8	A	U	K	U	K	K
9	B	U	K	K	U	K
10	A	K	U	U	U	K