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

Question

A particle moves such that its displacement $x$ varies with time $t$ as $x = At^2 + Bt^3$.  The graph of velocity versus time will be:

SUBHANI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
	**Primary Concept**: Kinematics and Calculus
	**Related Topics**: Differentiation, Velocity, Acceleration, Graphs of Motion
	**Prerequisites**: Basic calculus (differentiation), understanding of velocity as the derivative of displacement with respect to time.
## SOLUTION PATHWAY
**Given Information**:
	Displacement equation: $x = At^2 + Bt^3$

**Method & Steps**:
	1. Find the velocity by differentiating the displacement equation with respect to time.
		Reasoning: Velocity is the rate of change of displacement with respect to time, which is the first derivative of the displacement equation.
		Formula/Rule: $v = \frac{dx}{dt}$
	2. Perform the differentiation.
		Working: $v = \frac{d}{dt}(At^2 + Bt^3) = 2At + 3Bt^2$
		Key Point: The result is a quadratic equation in terms of time.
	3. Analyze the velocity equation to determine the shape of the graph.
		Result: The equation $v = 3Bt^2 + 2At$ is a quadratic equation of the form $y = ax^2 + bx + c$, where $a = 3B$, $b = 2A$ and $c = 0$. Since the coefficient of the $t^2$ term ($3B$) determines the direction of the parabola and is positive, the graph will be a parabola opening upwards.
		Verification: A positive coefficient for the $t^2$ term means the parabola will open upwards.
 ## OPTION ANALYSIS
[A] :x: A straight line indicates a constant acceleration, which is not the case here since the velocity is a quadratic function of time.
[B] :x: A straight line with a negative slope would mean a constant negative acceleration, not applicable in this case.
[C] ✓: The velocity is a quadratic function of time ($v = 3Bt^2 + 2At$) which represents a parabola. Since the coefficient of $t^2$ is positive the parabola opens upwards.
[D] :x: A parabola opening downwards would require a negative coefficient for the $t^2$ term in the velocity equation, which is not the case here.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Confusing displacement with velocity or acceleration. Remember that velocity is the first derivative of displacement and acceleration is the first derivative of velocity.
		2.[Error 2]: Not being able to interpret the quadratic equation for the velocity. Recognize that the highest power of 't' determines the curve shape on a graph.
	**Quick Tips**:
		1.[Helpful shortcut]: If the displacement is a polynomial of degree 'n', the velocity is a polynomial of degree 'n-1', and acceleration is a polynomial of degree 'n-2'.
		2.[Memory aid]: Remember that the derivative of $t^n$ is $nt^{n-1}$.

SUBHANI · Answer

A straight line with a positive slope.

SUBHANI · Answer

A straight line with a negative slope.

SUBHANI · Answer

The correct Option is "A parabola opening upwards."

SUBHANI · Answer

A parabola opening downwards.

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