Fisikastudycenter.com-The following shows some examples of questions
and discussion of the material covered in class XI Kinematics (11) High
School:
No Problem. 1
A particle moves with position versus time equation:
r (t) = 3t 2-2t +1
with t in the second and r in meters.
b. Average speed of the particle at t = 0 second to t = 2 second
No Problem. 2
An object moving straight at speed equation:
If the position of objects initially at the origin, then the displacement of the object during the 3 second is ...
A. 10 m
B. 20 m
C. 30 m
D. 40 m
E. 50 m
(Source question: Marthen Kanginan 2A, Kinematics with Vector Analysis)
Discussion
If the velocity equation is known, to find common positions integralkan the velocity equation, enter the requested time.
No Problem. 3
Graph velocity (v) versus time (t) following the motion of an object informs.
The average velocity of the initial movement of objects up to 18 seconds is ....
A. 3 m / s.
B. 6 m / s.
C. 9 m / s.
D. 12 m / s
E. 15 m / s
Discussion
The average velocity is the displacement divided by the time interval. If the provided graph v against t like about the above, the displacement can be found by searching the area under the curve by giving a positive sign if the above t-axis and the negative sign for under the axis t. Area = displacement = area of triangle + broad trapezoidal
No Problem. 4
Equation of angular position of an object moving in a circle is expressed as follows:
Define:
a) The starting position
b) The position at t = 2 second
c) the average angular velocity from t = 1 second to t = 2 second
d) initial angular velocity
e) Angular velocity at t = 1 second
f) The time when the particles stop moving
g) the average angular acceleration between t = 1 second to t = 2 second
h) Acceleration initial angle
i) Acceleration angle at t = 1 second
Discussion
a) The starting position is the position at t = 0 seconds, insert it into the position equation
b) The position at t = 2 second
c) the average angular velocity from t = 1 second to t = 2 second
d) initial angular velocity
Enter initial angular velocity t = 0 second at angular velocity equation. Because the equation is unknown lower angle position to obtain the angular velocity equation.
e) Angular velocity at t = 1 second
f) The time when the particles stop moving
Stop means zero angular velocity.
g) the average angular acceleration between t = 1 second to t = 2 second
h) Acceleration initial angle
Lower the angular velocity equation to get the equation of angular acceleration.
i) Acceleration angle at t = 1 second
No Problem. 5
A particle moves from the ground to the position equation Y = (-3t 2 + 12t + 6) feet. Define:
a) The initial position of the particle
b) The position of the particle at t = 1 second
c) The initial velocity of the particle
d) Acceleration of particles
e) The time it takes the particles to reach the highest point
f) Duration of the particles are in the air
g) The maximum height that can be achieved particle
Discussion
a) The initial position of the particle
b) The position of the particle at t = 1 second
c) The initial velocity of the particle
d) Acceleration of particles. Lower the speed equation to obtain the acceleration equation:
e) The time it takes the particles to reach the highest point
When it reaches the highest point of the particle velocity is zero.
f) Duration of the particles are in the air
Particles present in the air for twice the time to reach the highest point of the 4 second.
g) The maximum height that can be achieved particle
The maximum height is reached when 2 seconds, plug it into the position equation.
No Problem. 1
A particle moves with position versus time equation:
r (t) = 3t 2-2t +1
with t in the second and r in meters.
Define:
a. Particle velocity at t = 2 second
b. Average speed of the particle between t = 0 second to t = 2 second
Discussion
a. Particle velocity at t = 2 seconds (instantaneous velocity)
b. Average speed of the particle at t = 0 second to t = 2 second
No Problem. 2
An object moving straight at speed equation:
If the position of objects initially at the origin, then the displacement of the object during the 3 second is ...
A. 10 m
B. 20 m
C. 30 m
D. 40 m
E. 50 m
(Source question: Marthen Kanginan 2A, Kinematics with Vector Analysis)
Discussion
If the velocity equation is known, to find common positions integralkan the velocity equation, enter the requested time.
No Problem. 3
Graph velocity (v) versus time (t) following the motion of an object informs.
The average velocity of the initial movement of objects up to 18 seconds is ....
A. 3 m / s.
B. 6 m / s.
C. 9 m / s.
D. 12 m / s
E. 15 m / s
Discussion
The average velocity is the displacement divided by the time interval. If the provided graph v against t like about the above, the displacement can be found by searching the area under the curve by giving a positive sign if the above t-axis and the negative sign for under the axis t. Area = displacement = area of triangle + broad trapezoidal
No Problem. 4
Equation of angular position of an object moving in a circle is expressed as follows:
Define:
a) The starting position
b) The position at t = 2 second
c) the average angular velocity from t = 1 second to t = 2 second
d) initial angular velocity
e) Angular velocity at t = 1 second
f) The time when the particles stop moving
g) the average angular acceleration between t = 1 second to t = 2 second
h) Acceleration initial angle
i) Acceleration angle at t = 1 second
Discussion
a) The starting position is the position at t = 0 seconds, insert it into the position equation
b) The position at t = 2 second
c) the average angular velocity from t = 1 second to t = 2 second
d) initial angular velocity
Enter initial angular velocity t = 0 second at angular velocity equation. Because the equation is unknown lower angle position to obtain the angular velocity equation.
e) Angular velocity at t = 1 second
f) The time when the particles stop moving
Stop means zero angular velocity.
g) the average angular acceleration between t = 1 second to t = 2 second
h) Acceleration initial angle
Lower the angular velocity equation to get the equation of angular acceleration.
i) Acceleration angle at t = 1 second
No Problem. 5
A particle moves from the ground to the position equation Y = (-3t 2 + 12t + 6) feet. Define:
a) The initial position of the particle
b) The position of the particle at t = 1 second
c) The initial velocity of the particle
d) Acceleration of particles
e) The time it takes the particles to reach the highest point
f) Duration of the particles are in the air
g) The maximum height that can be achieved particle
Discussion
a) The initial position of the particle
b) The position of the particle at t = 1 second
c) The initial velocity of the particle
d) Acceleration of particles. Lower the speed equation to obtain the acceleration equation:
e) The time it takes the particles to reach the highest point
When it reaches the highest point of the particle velocity is zero.
f) Duration of the particles are in the air
Particles present in the air for twice the time to reach the highest point of the 4 second.
g) The maximum height that can be achieved particle
The maximum height is reached when 2 seconds, plug it into the position equation.
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