The Question is from neet-physics motion-in-a-straight-line distance-and-displacement, and it was created by the author SUBHANI

Question

A particle moves along a trajectory defined by $x = t^3 - 6t^2 + 9t + 2$ and $y = t^2 - 2t + 1$, where $x$ and $y$ are in meters and $t$ is in seconds.  What is the magnitude of the displacement of the particle between $t=0$ and $t=4$ seconds?

SUBHANI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Displacement Calculation
	**Related Topics**: Kinematics, Vector Analysis
	**Prerequisites**: Position, Displacement, Vector Magnitude
## SOLUTION PATHWAY
**Given Information**:
	Initial time: $t_1 = 0$ s
	Final time: $t_2 = 4$ s
	Position function x(t) = $t^3 - 6t^2 + 9t + 2$
	Position function y(t) = $t^2 - 2t + 1$
**Method & Steps**:
	1.	Find the initial position of the particle at $t=0$.
		Reasoning: Substitute $t=0$ into the position equations for x and y to find the starting coordinates.
		Formula/Rule: $x(t)$ and $y(t)$ evaluated at $t=0$
	2.	Calculate $x_1$ and $y_1$ at $t=0$.
		Working: $x(0) = (0)^3 - 6(0)^2 + 9(0) + 2 = 2$
		Working: $y(0) = (0)^2 - 2(0) + 1 = 1$
		Key Point: Initial position is $(x_1, y_1) = (2, 1)$
	3.	Find the final position of the particle at $t=4$.
		Reasoning: Substitute $t=4$ into the position equations for x and y to find the ending coordinates.
		Formula/Rule: $x(t)$ and $y(t)$ evaluated at $t=4$
	4.	Calculate $x_2$ and $y_2$ at $t=4$.
		Working: $x(4) = (4)^3 - 6(4)^2 + 9(4) + 2 = 64 - 96 + 36 + 2 = 6$
		Working: $y(4) = (4)^2 - 2(4) + 1 = 16 - 8 + 1 = 9$
		Key Point: Final position is $(x_2, y_2) = (6, 9)$
	5.	Calculate the displacement vector.
		Reasoning: The displacement vector is the difference between final and initial position vectors.
		Formula/Rule: $\Delta \vec{r} = (x_2 - x_1) \hat{i} + (y_2 - y_1) \hat{j}$
	6.	Calculate the displacement vector components.
		Working: $\Delta x = 6 - 2 = 4$
		Working: $\Delta y = 9 - 1 = 8$
		Key Point: The displacement vector is $\Delta \vec{r} = 4\hat{i} + 8\hat{j}$
	7.	Calculate the magnitude of the displacement vector.
		Reasoning: The magnitude of a vector is the square root of the sum of the squares of its components.
		Formula/Rule:  $|\Delta \vec{r}| = \sqrt{(\Delta x)^2 + (\Delta y)^2}$
	8.	Calculate the magnitude of displacement.
		Working: $|\Delta \vec{r}| = \sqrt{(4)^2 + (8)^2} = \sqrt{16 + 64} = \sqrt{80} = \sqrt{16 	imes 5} = 4\sqrt{5}$
		Result: The magnitude of the displacement is $4\sqrt{5}$ m
		Verification: The calculation is correct.
 ## OPTION ANALYSIS
[A] :x: Incorrect magnitude of the displacement.
[B] ✓: Correct magnitude of the displacement.
[C] :x: Incorrect magnitude of the displacement.
[D] :x: Incorrect magnitude of the displacement.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Incorrectly calculating the position at t=0 or t=4. : [Double check the substitutions into x(t) and y(t)]
		2.[Error 2]: Incorrectly calculating the magnitude of the displacement vector. : [Remember to square the components before adding them and then taking the square root]
	**Quick Tips**:
		1.[Helpful shortcut]: Always use the final position minus initial position for displacement calculations.
		2.[Memory aid]: Displacement is a vector, so consider both x and y components.

SUBHANI · Answer

2*sqrt(5)

SUBHANI · Answer

The correct Option is "4*sqrt(5)"

SUBHANI · Answer

6*sqrt(5)

SUBHANI · Answer

8*sqrt(5)

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