Kinematics
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Kinematics: an area in physics that analyze that focuses and analyze the motion in terms of time and distance
In badminton, the most important skill that you must know and understand is smashing and clearing. Smashing is a key component in badminton. If you don't understand the concept of smashing or can't smash, you basically have lost the game. Smashing is a key example of high velocity/high speed. One of the fastest smashes was achieved by Naoki Kawamae. He covered an area of 9m which took a total of 0.0782 seconds. To figure out the speed achieved, we can use the formula for speed(velocity) which is v = d/t, 'v' which represents the speed/velocity, 'd' represents the distance and 't' represents the time taken to cover that distance. In this example v = 9m/0.0782s which then v = 115m/s which is then translated to 414km/h. Speed is referred to as scalar because it just a term of measurement. Velocity is a speed also but it is a vector because it has a direction in where to goes. Acceleration is the change in speed of an object over a certain period of time.
In badminton, the most important skill that you must know and understand is smashing and clearing. Smashing is a key component in badminton. If you don't understand the concept of smashing or can't smash, you basically have lost the game. Smashing is a key example of high velocity/high speed. One of the fastest smashes was achieved by Naoki Kawamae. He covered an area of 9m which took a total of 0.0782 seconds. To figure out the speed achieved, we can use the formula for speed(velocity) which is v = d/t, 'v' which represents the speed/velocity, 'd' represents the distance and 't' represents the time taken to cover that distance. In this example v = 9m/0.0782s which then v = 115m/s which is then translated to 414km/h. Speed is referred to as scalar because it just a term of measurement. Velocity is a speed also but it is a vector because it has a direction in where to goes. Acceleration is the change in speed of an object over a certain period of time.
In every game of badminton, every contact between your racket and the shuttlecock may be different. But speed is the key characteristic. In badminton, like many other sports, the velocity of an object is not always consistent and may vary. Due to this, there will either an acceleration or a deceleration. For example, suppose you hit a shuttlecock, it travels 16m with a velocity of 9m/s. What is the acceleration of the shuttlecock? To answer this, it can be calculated using one of the five equations of linear kinematics.
V is the velocity, d is the distance(m), a is the acceleration(m/s^2), and t is the time taken(s)
Equations of Linear Kinematics
V2=V1+aΔt
Δd=1/2(V2+V1)Δt
Δd=V1Δt+(1/2)aΔt^2
Δd=V2Δt-(1/2)aΔt^2
V2^2=V1^2+2aΔd
Using the 5 Equation of Linear Kinematics we can solve for the acceleration of covering 16m in 9m/s
Δd = 16m
V1 = 0
V2 = 9m/s
a = ?
V2^2=V1^2+2aΔd
9^2 = 0^2 + 2(a)(16)
81 = 0 + 32a
a = 2.5m/s^2
∴The acceleration of the shuttlecock is 2.5m/s^2
Since we live on Earth, there is gravity. The rule is, what goes up, must come down. For example, in space if an object is thrown it will travel in a straight line until it hits another object or a type of gravity comes into play. But if the same object is thrown on Earth, it creates an arc/parabolic motion instead of a continuous straight line. This specific motion is referred to as projectile motion. In badminton, like many other sports, when an object is thrown it creates a arc. For example, a shuttlecock is hit with a horizontal velocity of 9m/s and a vertical velocity of 3.7m/s, and the shuttlecock was hit from a height of 1.6m. How far did it travel? To solve this, layout all known variables in horizontal and vertical components.
Horizontal
V1h = V2h = 9m/s
Ah = 0
ΔDh = ?
Δt = ?
Vertical
V1v = 3.7m/s
Av = -9.8m/s^2
ΔDv = 1.6m
Δt= ?
Since we know that the time in both components will be the same, we can use the vertical components and substitute into the horizontal components.
Equation 3 from the linear kinetics suits all the given variables.
ΔDv = V1vΔt + (1/2)AvΔt^2
1.6 = 3.7Δt + (1/2)(-9.8)Δt^2
0 = -4.9Δt^2 + 3.7Δt - 1.6
To solve for t, must substitute above answer into Quadratic Formula.
x = [-b ± √(b^2 - 4ac)] /2a
t = [-3.7 ± √(3.7^2) - 4(-4.9)(-1.6)] /2(-4.9)
t = 0.8064 or t = -0.0513
∴The time taken is 0.81 seconds
Once that the time taken is solved which is 0.81 seconds, substitute it into the horizontal components and find the distance. Equation 3 still remains as the best equation since all variables solved now are given in the equation.
ΔDh = V1hΔt + (1/2)AhΔt^2
ΔDh = 9(0.8064) + (1/2)(0)(0.8064)
ΔDh = 7.25m
The shuttlecock traveled a total distance of 7.25m
V is the velocity, d is the distance(m), a is the acceleration(m/s^2), and t is the time taken(s)
Equations of Linear Kinematics
V2=V1+aΔt
Δd=1/2(V2+V1)Δt
Δd=V1Δt+(1/2)aΔt^2
Δd=V2Δt-(1/2)aΔt^2
V2^2=V1^2+2aΔd
Using the 5 Equation of Linear Kinematics we can solve for the acceleration of covering 16m in 9m/s
Δd = 16m
V1 = 0
V2 = 9m/s
a = ?
V2^2=V1^2+2aΔd
9^2 = 0^2 + 2(a)(16)
81 = 0 + 32a
a = 2.5m/s^2
∴The acceleration of the shuttlecock is 2.5m/s^2
Since we live on Earth, there is gravity. The rule is, what goes up, must come down. For example, in space if an object is thrown it will travel in a straight line until it hits another object or a type of gravity comes into play. But if the same object is thrown on Earth, it creates an arc/parabolic motion instead of a continuous straight line. This specific motion is referred to as projectile motion. In badminton, like many other sports, when an object is thrown it creates a arc. For example, a shuttlecock is hit with a horizontal velocity of 9m/s and a vertical velocity of 3.7m/s, and the shuttlecock was hit from a height of 1.6m. How far did it travel? To solve this, layout all known variables in horizontal and vertical components.
Horizontal
V1h = V2h = 9m/s
Ah = 0
ΔDh = ?
Δt = ?
Vertical
V1v = 3.7m/s
Av = -9.8m/s^2
ΔDv = 1.6m
Δt= ?
Since we know that the time in both components will be the same, we can use the vertical components and substitute into the horizontal components.
Equation 3 from the linear kinetics suits all the given variables.
ΔDv = V1vΔt + (1/2)AvΔt^2
1.6 = 3.7Δt + (1/2)(-9.8)Δt^2
0 = -4.9Δt^2 + 3.7Δt - 1.6
To solve for t, must substitute above answer into Quadratic Formula.
x = [-b ± √(b^2 - 4ac)] /2a
t = [-3.7 ± √(3.7^2) - 4(-4.9)(-1.6)] /2(-4.9)
t = 0.8064 or t = -0.0513
∴The time taken is 0.81 seconds
Once that the time taken is solved which is 0.81 seconds, substitute it into the horizontal components and find the distance. Equation 3 still remains as the best equation since all variables solved now are given in the equation.
ΔDh = V1hΔt + (1/2)AhΔt^2
ΔDh = 9(0.8064) + (1/2)(0)(0.8064)
ΔDh = 7.25m
The shuttlecock traveled a total distance of 7.25m