Kinematics is a subdivision of mechanics which describes the motion of objects in one, two, and three dimensional spaces without reference to force or mass. Kinematics explains the effect of motion, not the causes of motion. It involves the position, velocity, and acceleration of objects.
Explain:
An object rolls down a ramp 10 meters tall and 15 meters in length with an acceleration of 4.9m/s/s. The initial velocity is 0. The object's velocity is at it's maximum at the bottom of the ramp. Every second, the velocity increases and the object changes position. When the object reaches the bottom of the ramp, it's displacement is 18 meters. sqrt((10^2)+(15^2))
Examples:
Starting from rest, a rocket accelerates at 29.4 m/s/s. The rocket stops accelerating when it reaches a velocity of 500 m/s, then drifts with a constant velocity for 200 seconds. How far did the rocket travel?
Solve for the time that the rocket is accelerating.
Vf = Vi + a * t
500m/s = 0 + 29.4m/s/s * t
500 / 29.4 = t
t = 17s
The rocket accelerates at 29.4 m/s/s for 17s.
With the time and acceleration, you can solve for the displacement.
d = 1/2at²
d = 1/2(29.4)17²
d = 14.7 * 289
d = 4248.3m
The rocket travels 4248.3m during the time it accelerates.
Solve for the distance traveled in the next 200 seconds.
v = d / t
500m/s = d / 200
d = 100,000m
The rocket travels 100,000m in 200s at a constant velocity of 500m/s 4248.3m + 100,000m = 104248.3m
The rocket travels a total of 104248.3m.
Equations:
d = distance
x = position
Vi = initial velocity
Vf = final velocity
t = time
a = acceleration
-
d = 1/2at²
X = Xi + Vi * t + 1/2at²
Vf = Vi + a * t
d = ((Vi + Vf) / 2) * t
Vf² = Vi² + 2ad
Kinematics
Define:
Kinematics is a subdivision of mechanics which describes the motion of objects in one, two, and three dimensional spaces without reference to force or mass. Kinematics explains the effect of motion, not the causes of motion. It involves the position, velocity, and acceleration of objects.Explain:
An object rolls down a ramp 10 meters tall and 15 meters in length with an acceleration of 4.9m/s/s. The initial velocity is 0. The object's velocity is at it's maximum at the bottom of the ramp. Every second, the velocity increases and the object changes position. When the object reaches the bottom of the ramp, it's displacement is 18 meters. sqrt((10^2)+(15^2))
Examples:
Starting from rest, a rocket accelerates at 29.4 m/s/s. The rocket stops accelerating when it reaches a velocity of 500 m/s, then drifts with a constant velocity for 200 seconds. How far did the rocket travel?Solve for the time that the rocket is accelerating.
Vf = Vi + a * t
500m/s = 0 + 29.4m/s/s * t
500 / 29.4 = t
t = 17s
The rocket accelerates at 29.4 m/s/s for 17s.
With the time and acceleration, you can solve for the displacement.
d = 1/2at²
d = 1/2(29.4)17²
d = 14.7 * 289
d = 4248.3m
The rocket travels 4248.3m during the time it accelerates.
Solve for the distance traveled in the next 200 seconds.
v = d / t
500m/s = d / 200
d = 100,000m
The rocket travels 100,000m in 200s at a constant velocity of 500m/s
4248.3m + 100,000m = 104248.3m
The rocket travels a total of 104248.3m.
Equations:
d = distancex = position
Vi = initial velocity
Vf = final velocity
t = time
a = acceleration
-
d = 1/2at²
X = Xi + Vi * t + 1/2at²
Vf = Vi + a * t
d = ((Vi + Vf) / 2) * t
Vf² = Vi² + 2ad
Other topics
Sources
http://www.physics-online.info/book1/chapt1/Topic1-4.htmhttp://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/1DKin/U1L6a.html
http://www.physicsclassroom.com/Class/1DKin/U1L6e.cfm