Kinematics – One Direction

Kinematics is the study of motion without consideration of forces. There are four fundamental equations for movement in one direction. These encompass the vector quantities previously discussed – displacement, velocity and acceleration.

kinematics

Equation 1: d = vi * t + ½ * a * t2

Here, d is displacement, vi is initial velocity, a is acceleration and t is time.

This equation can be used to determine the displacement of a moving object with known initial velocity and acceleration and time. Furthermore, any of the above quantities can be determined if the other are present.

Equation 2: vf2 = vi2 + 2 * a * d

This equation is usually used to determine either the final velocity or rearranged to find the displacement.

Equation 3: vf =vi + a * t

Final velocity is equal to the sum of initial velocity and acceleration multiplied by time taken.

This can be used to determine the final velocity. This equation is typically rearranged to find the acceleration.

Equation 4: d = (vi + vf ) / 2 * t

This equation calculates the average velocity and multiplies it with time to give us the displacement.

Kinematics can either be solved in the X or Y direction, depending on the problem.

Special Cases:

If the problem deals with Y-direction movement,

  • Acceleration = g = -9.81 m/s2
  • If the ball is vertically thrown in the air and then falls down to the ground,
    • Initial velocity (velocity at beginning of launch) = final velocity (velocity right before reaching the ground)
    • The motion can be split in half, so the first half of the motion (when it reaches the highest point) is the same as the second half (from the highest point to the ground)
    • The velocity at the highest point is zero. At the highest point, the object momentarily stops. Hence, the velocity is zero at that point.

Recommended Steps:

1. Draw diagram of situation. This will help you visualize the initial and final position as well as the direction of the moving object.
2. Write the variables given. Make assumptions based on the details of the problem.
3. List the variable that must be solved.

4. Find an equation(s) that relates all the variables together.
a. An equation must be selected in which all, but one variable must be known. If you pick an equation in which 2 or more of the variables are unknown, then a different equation must be picked.

5. Solve for the missing variable.
a. Tip: Isolate the variable that needs to be solved first, then substitute the known values.

Example: Michael is waiting at a stoplight. When it finally turns green, Michael accelerates from rest at a rate of 6.00 m/s2 for a time of 4.10 seconds. Determine the displacement of Michael’s car during this time period.

one direction

Given, vi = 0 m/s t= 4.10s a = 6 m/s2

Use equation 1: d = vi * t + ½ * a * t2

d = 0 * 4.10 + ½ * 6 * 4.10^2
= 0 + 3* 16.81
= 50.43 m

Practice Problems

1. An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until it finally lifts off the ground. Determine the distance traveled before takeoff.

Solution:

1. Write variables given. Make assumptions.

Given:

a = 3.20 m/s2
t = 32.8 sec
vi = 0 (object starts from rest)

Assumptions:

Starting position = 0 m (origin)

2. Write the variable that needs to be solved.

Solve for: d

3. Find an equation(s) that relates variables together.

Use Equation 1: d = vi * t + ½ * a * t2

4. Substitute variables in equation and solve for missing variable.

d = vi * t + ½ * a * t2
d = 0 * (32.8) + ½ * (3.20) * (32.8)2
d = 1721.344 m

2. A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.

Solution:

1. Write variables given. Make assumptions.

Given:

t = 5.21 sec
vi = 0 (object starts from rest)
d = 110 m

Assumptions:

Starting position = 0 m (origin)

2. Write the variable that needs to be solved

Solve for: a

3. Find an equation(s) that relates variables together.

Use Equation 1: d = vi * t + ½ * a * t2

4. Substitute variables in equation and solve for missing variable.

d = vi * t + ½ * a * t2
110 = 0 * 5.21 + ½ * (a)* (5.21)2
110 = 13.5720 (a)
a 8.104 m/s2

3. A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled.

Solution:

1. Write variables given. Make assumptions.

Given:

t = 2.47 sec
vi = 18.5 m/s
vf = 46.1 m/s

Assumptions:

Starting position = 0 m (origin)

2. Write variable that needs to be solved.

Solve for:
a
d

3. Find an equation(s) that relates the variables together. Substitute known variables and solve for missing variable.

Equation 3: a = (Vf – Vi)/ time = (46.1 – 18.5) / 2.47 = 11.174 m/s2
Equation 4: d = (vi + vf ) / 2 * t

d = (vi + vf ) / 2 * t
d = (46.1 + 18.5)/2 * 2.47
d = 79.781 m

Reference

https://www.physicsclassroom.com/class/1DKin/Lesson-6/Kinematic-Equations-and-Free-Fall