Question:
The coordinates of a particle moving in the \( x \)-\( y \) plane are given by: \[x = 2 + 4t, \quad y = 3t + 8t^2.\]The motion of the particle is:
The coordinates of a particle moving in the \( x \)-\( y \) plane are given by: \[x = 2 + 4t, \quad y = 3t + 8t^2.\]The motion of the particle is:
Updated On: Mar 19, 2026
- non-uniformly accelerated.
- uniformly accelerated having motion along a straight line.
- uniform motion along a straight line.
- uniformly accelerated having motion along a parabolic path.
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The Correct Option is D
Approach Solution - 1
Calculate the Velocity Components:
For \(x = 2 + 4t\):
\[ \frac{dx}{dt} = v_x = 4 \]For \(y = 3t + 8t^2\):
\[ \frac{dy}{dt} = v_y = 3 + 16t \]Calculate the Acceleration Components:
The acceleration in the \(x\)-direction \(a_x\) is:
\[ \frac{d^2x}{dt^2} = a_x = 0 \]The acceleration in the \(y\)-direction \(a_y\) is:
\[ \frac{d^2y}{dt^2} = a_y = 16 \]Therefore, the particle has a constant acceleration \(a_y = 16\, \text{m/s}^2\) in the \(y\)-direction, and no acceleration in the \(x\)-direction.
Determine the Path of Motion:
To find the path of the particle, express \(y\) in terms of \(x\) by eliminating \(t\) between the two equations.
From \(x = 2 + 4t\), we get:
\[ t = \frac{x - 2}{4} \]Substitute this into the equation for \(y\):
\[ y = 3\left(\frac{x - 2}{4}\right) + 8\left(\frac{x - 2}{4}\right)^2 \]Simplifying, we get:
\[ y = \frac{3}{4}(x - 2) + 8 \times \frac{(x - 2)^2}{16} \] \[ y = \frac{3}{4}(x - 2) + \frac{1}{2}(x - 2)^2 \]This equation is quadratic in \(x\), indicating that the path of the particle is a parabola.
Conclusion:
The motion of the particle is uniformly accelerated with a parabolic path, which corresponds to Option (4).
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Approach Solution -2
Step 1: Given equations of motion.
The position of a particle moving in the x–y plane is given by:
\[ x = 2 + 4t, \quad y = 3t + 8t^2 \] We need to determine the nature of the motion.
Step 2: Determine velocity components.
Differentiate each coordinate with respect to time \( t \):
\[ v_x = \frac{dx}{dt} = 4 \] \[ v_y = \frac{dy}{dt} = 3 + 16t \] Here, \( v_x \) is constant, while \( v_y \) increases linearly with time.
Step 3: Determine acceleration.
Differentiate again with respect to time:
\[ a_x = \frac{dv_x}{dt} = 0, \quad a_y = \frac{dv_y}{dt} = 16 \] Hence, acceleration is constant and directed along the y-axis. Therefore, the motion is uniformly accelerated.
Step 4: Determine the path.
Eliminate \( t \) between the two equations.
From \( x = 2 + 4t \), we get \( t = \frac{x - 2}{4} \).
Substitute this into \( y = 3t + 8t^2 \):
\[ y = 3\left(\frac{x - 2}{4}\right) + 8\left(\frac{x - 2}{4}\right)^2 \] \[ y = \frac{3(x - 2)}{4} + \frac{8(x - 2)^2}{16} = \frac{3(x - 2)}{4} + \frac{(x - 2)^2}{2} \] This is a quadratic equation in \( x \), which represents a parabola.
Step 5: Conclusion.
The particle moves along a parabolic path with a constant acceleration.
Final Answer:
\[ \boxed{\text{Uniformly accelerated motion along a parabolic path.}} \]
The position of a particle moving in the x–y plane is given by:
\[ x = 2 + 4t, \quad y = 3t + 8t^2 \] We need to determine the nature of the motion.
Step 2: Determine velocity components.
Differentiate each coordinate with respect to time \( t \):
\[ v_x = \frac{dx}{dt} = 4 \] \[ v_y = \frac{dy}{dt} = 3 + 16t \] Here, \( v_x \) is constant, while \( v_y \) increases linearly with time.
Step 3: Determine acceleration.
Differentiate again with respect to time:
\[ a_x = \frac{dv_x}{dt} = 0, \quad a_y = \frac{dv_y}{dt} = 16 \] Hence, acceleration is constant and directed along the y-axis. Therefore, the motion is uniformly accelerated.
Step 4: Determine the path.
Eliminate \( t \) between the two equations.
From \( x = 2 + 4t \), we get \( t = \frac{x - 2}{4} \).
Substitute this into \( y = 3t + 8t^2 \):
\[ y = 3\left(\frac{x - 2}{4}\right) + 8\left(\frac{x - 2}{4}\right)^2 \] \[ y = \frac{3(x - 2)}{4} + \frac{8(x - 2)^2}{16} = \frac{3(x - 2)}{4} + \frac{(x - 2)^2}{2} \] This is a quadratic equation in \( x \), which represents a parabola.
Step 5: Conclusion.
The particle moves along a parabolic path with a constant acceleration.
Final Answer:
\[ \boxed{\text{Uniformly accelerated motion along a parabolic path.}} \]
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