An Object Has A Position Given By R = [2.0 M + (5.00 M/s)t] I^ + [3.0 M (3.00 M/s2)t2] J^ , Where Quantities
Understanding the position of an object in physics is fundamental to analyzing its motion. The given position vector
\[ R = [2.0\, \text{M} + (5.00\, \text{M/s})t]\, \hat{\imath} + [3.0\, \text{M} \times (3.00\, \text{M/s}^2)t^2]\, \hat{\jmath} \]
provides a comprehensive way to describe the object's location at any given time \( t \). This article explores the various quantities involved, how to interpret this position vector, and the underlying physics concepts such as velocity, acceleration, and trajectory that can be derived from this expression.
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Deciphering the Position Vector R
The position vector \( R \) details the object’s position in a two-dimensional plane, combining the components along the x-axis (\( \hat{\imath} \)) and y-axis (\( \hat{\jmath} \)).
Component Breakdown
The vector can be split into two main parts:- X-component: \( R_x(t) = 2.0\, \text{M} + (5.00\, \text{M/s}) t \)
- Y-component: \( R_y(t) = 3.0\, \text{M} \times (3.00\, \text{M/s}^2) t^2 \)
This indicates:
- The x-position starts at 2.0 meters when \( t = 0 \) and increases linearly with time at a rate of 5.00 meters per second.
- The y-position starts at zero (since \( t = 0 \), \( R_y = 0 \)), but accelerates quadratically over time, following a parabolic trajectory.
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Interpreting the Quantities Involved
Understanding the physical meaning of each term and quantity is essential for analyzing the object's motion.
Initial Position Components
- Initial x-position (\( R_{x0} \)): 2.0 meters
- Initial y-position (\( R{y0} \)): 0 meters (since \( t = 0 \), \( Ry = 0 \))
Velocity Components
- The x-velocity component is constant at 5.00 m/s, indicative of uniform motion in the x-direction.
- The y-velocity component is time-dependent because of acceleration, as it involves a quadratic term.
Quantities and Their Physical Significance
- Position (\( R \)): The location of the object at a specific time \( t \).
- Time (\( t \)): The independent variable determining the position.
- Velocity (\( v \)): The rate of change of position with respect to time, which can be obtained by differentiating \( R \).
- Acceleration (\( a \)): The rate of change of velocity, especially relevant in the y-direction where quadratic dependence indicates acceleration.
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Calculating Velocity from the Position Vector
Velocity is a crucial quantity in kinematics, describing how fast and in what direction an object moves.
Deriving Velocity Components
To find the velocity components, differentiate each component of \( R \) with respect to time:\[
vx(t) = \frac{d Rx}{dt} = \frac{d}{dt} [2.0\, \text{M} + 5.00\, \text{M/s} \times t] = 5.00\, \text{M/s}
\]
\[
vy(t) = \frac{d Ry}{dt} = \frac{d}{dt} [3.0\, \text{M} \times 3.00\, \text{M/s}^2 \times t^2] = 3.0\, \text{M} \times 3.00\, \text{M/s}^2 \times 2t = 18.0\, \text{M}^2/\text{s}^2 \times t
\]
Note that:
- The x-velocity remains constant at 5.00 m/s.
- The y-velocity varies linearly with time, indicating acceleration in the y-direction.
Summary of Velocity Components
- \( v_x(t) = 5.00\, \text{M/s} \)
- \( v_y(t) = 18.0\, \text{M}^2/\text{s}^2 \times t \)
Implications
- The object moves uniformly along the x-axis.
- The y-motion involves acceleration, with the velocity increasing linearly over time, typical of uniformly accelerated motion.
Calculating Acceleration
Acceleration describes how the velocity changes over time.
Deriving Acceleration Components
Differentiate the velocity components:\[
ax(t) = \frac{d vx}{dt} = 0
\]
\[
ay(t) = \frac{d vy}{dt} = 18.0\, \text{M}^2/\text{s}^2
\]
This indicates:
- No acceleration in the x-direction (\( a_x = 0 \)), consistent with constant velocity.
- Constant acceleration in the y-direction (\( a_y = 18.0\, \text{M}^2/\text{s}^2 \)).
Physical Significance
- The object experiences uniform acceleration vertically, with magnitude 18.0 m/s².
- The acceleration causes the quadratic increase in the y-position over time.
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Analyzing the Trajectory of the Object
The path traced by the object can be visualized by plotting \( Rx(t) \) versus \( Ry(t) \).
Equation of the Trajectory
Express \( Ry \) in terms of \( Rx \):\[
R_x(t) = 2.0\, \text{M} + 5.00\, \text{M/s} \times t
\]
\[
t = \frac{R_x - 2.0\, \text{M}}{5.00\, \text{M/s}}
\]
Substitute into \( R_y(t) \):
\[
Ry = 3.0\, \text{M} \times 3.00\, \text{M/s}^2 \times \left( \frac{Rx - 2.0\, \text{M}}{5.00\, \text{M/s}} \right)^2
\]
Simplify:
\[
Ry = 3.0\, \text{M} \times 3.00\, \text{M/s}^2 \times \frac{(Rx - 2.0\, \text{M})^2}{(5.00\, \text{M/s})^2}
\]
\[
Ry = \frac{3.0 \times 3.00}{25.00} \times (Rx - 2.0)^2\, \text{M}
\]
\[
Ry = \frac{9.00}{25.00} \times (Rx - 2.0)^2\, \text{M} \approx 0.36 \times (R_x - 2.0)^2\, \text{M}
\]
This quadratic relation describes a parabolic trajectory, characteristic of projectile motion under constant acceleration.
Visual Interpretation
- The object starts at \( R = (2.0\, \text{M}, 0) \).
- As \( Rx \) increases linearly, \( Ry \) increases quadratically, forming a parabola.
Key Physical Quantities and Their Calculations at Specific Times
Knowing the position, velocity, and acceleration at specific moments can provide deeper insights into the motion.
At \( t = 0\, \text{s} \)
- \( R_x = 2.0\, \text{M} \)
- \( R_y = 0\, \text{M} \)
- \( v_x = 5.00\, \text{M/s} \)
- \( v_y = 0\, \text{M/s} \)
- \( a_x = 0\, \text{M/s}^2 \)
- \( a_y = 18.0\, \text{M/s}^2 \)
At \( t = 2\, \text{s} \)
- \( R_x
