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Question

The velocity-time graph of a particle is given by $v = v_0(1-e^{-kt})$, where $v_0$ and $k$ are constants. The displacement of the particle at time $t$ is:

ABDUL JILANI BASHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: Option C
## CONCEPT FRAMEWORK
	**Primary Concept**: Kinematics - Displacement from Velocity
	**Related Topics**: Integration, Exponential Functions
	**Prerequisites**: Basic calculus, understanding of velocity and displacement relationship
## SOLUTION PATHWAY
**Given Information**:
	* Velocity function: $v = v_0(1-e^{-kt})$
	*  Displacement is required at time t
**Method & Steps**:
	1. Find the displacement by integrating the velocity with respect to time.
		Reasoning: The displacement is the integral of velocity with respect to time, i.e., $s = \int v dt$.
		Formula/Rule: $s(t) = \int v(t) dt$
	2. Integrate the given velocity function.
		Working: $s(t) = \int v_0(1-e^{-kt}) dt = v_0 \int (1 - e^{-kt}) dt = v_0 [\int 1 dt - \int e^{-kt} dt] = v_0 [t - \frac{e^{-kt}}{-k}] = v_0 [t + \frac{e^{-kt}}{k}] + C$
		Key Point: Remember to add the constant of integration (C).
	3. Determine the constant of integration, assuming initial displacement at t=0 is 0. 
		Working: if s(0) = 0, then 0 = $v_0 [0 + \frac{e^{0}}{k}] + C$ => $0 = v_0(\frac{1}{k}) + C$ => $C = -\frac{v_0}{k}$
	4. Substitute value of C in displacement equation and simplify.
		Working: $s(t) = v_0 [t + \frac{e^{-kt}}{k}] - \frac{v_0}{k} = v_0 [t + \frac{e^{-kt}}{k} - \frac{1}{k}] = v_0[t + \frac{e^{-kt}-1}{k}]$
	5. Final result.
		Result: $s(t) = v_0[t + \frac{e^{-kt}-1}{k}]$
		Verification: The result is consistent with integration of the given velocity function and appropriate initial condition.
 ## OPTION ANALYSIS
[A] :x: This is the displacement if the velocity was constant at $v_0$. It does not consider the exponential term.
[B] :x:  This is dimensionally incorrect and doesnt account for time dependence.
[C] ✓: This is the correct displacement, obtained by integrating the given velocity function and applying the initial condition of zero displacement at t=0.
[D] :x: The sign within the bracket is incorrect. The integral of $e^{-kt}$ is $-e^{-kt}/k$ which when used in the displacement calculation leads to a positive $e^{-kt}/k$ term
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Forgetting to add the constant of integration, which can lead to an incorrect result if initial condition isn't zero
		2.[Error 2]: Incorrect integration of the exponential term.
	**Quick Tips**:
		1.[Helpful shortcut]: Remember that displacement is the integral of velocity with respect to time.
		2.[Memory aid]: The integral of $e^{ax}$ is $e^{ax}/a$.

ABDUL JILANI BASHA · Answer

$v_0t$

ABDUL JILANI BASHA · Answer

$v_0/k$

ABDUL JILANI BASHA · Answer

The correct Option is "$v_0[t + (e^{-kt}-1)/k]$"

ABDUL JILANI BASHA · Answer

$v_0[t - (e^{-kt}-1)/k]$

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