set 11k

Q1 [3 Marks]: Can a position-time graph ever be a straight line parallel to the position axis? Justify your answer physically.

Answer: NO.

Physical Justification:
The position axis is the vertical axis ($y$-axis), and the time axis is the horizontal axis ($x$-axis). A straight line parallel to the position axis would mean that at one single, exact instant in time, the object exists at an infinite number of different positions simultaneously.

Mathematically, the slope of a position-time graph gives velocity ($v = \frac{\Delta x}{\Delta t}$). A vertical line has an undefined or infinite slope ($\Delta t = 0$), which would imply the object is traveling at infinite velocity. Since no object can travel faster than the speed of light, nor can it be in multiple places at the exact same time, this graph is physically impossible.

Q2 [3 Marks]: A car accelerates from rest at a constant rate $\alpha$ for some time, after which it decelerates at a constant rate $\beta$ to come to rest. If the total time elapsed is $t$, evaluate the maximum velocity acquired.

Step 1: Set up the time intervals.
Let $t_1$ be the time spent accelerating, and $t_2$ be the time spent decelerating.

$$\text{Total time, } t = t_1 + t_2 \quad \text{--- (Equation 1)}$$

Step 2: Analyze the acceleration phase.
The car starts from rest ($u=0$), accelerates at $\alpha$, and reaches a maximum velocity $v_{max}$ at the end of time $t_1$.

$$v_{max} = u + \alpha t_1 = 0 + \alpha t_1$$

$$t_1 = \frac{v_{max}}{\alpha} \quad \text{--- (Equation 2)}$$

Step 3: Analyze the deceleration phase.
The car starts this phase at $v_{max}$, decelerates at $\beta$ (so acceleration is $-\beta$), and comes to a complete stop ($v=0$) after time $t_2$.

$$0 = v_{max} - \beta t_2$$

$$t_2 = \frac{v_{max}}{\beta} \quad \text{--- (Equation 3)}$$

Step 4: Substitute and solve for $v_{max}$.
Substitute Equations 2 and 3 into Equation 1:

$$t = \frac{v_{max}}{\alpha} + \frac{v_{max}}{\beta}$$

Factor out $v_{max}$:

$$t = v_{max} \left( \frac{1}{\alpha} + \frac{1}{\beta} \right) = v_{max} \left( \frac{\alpha + \beta}{\alpha \beta} \right)$$

Rearranging to isolate $v_{max}$ yields:

$$v_{max} = \frac{\alpha \beta}{\alpha + \beta} t$$

Q3 [4 Marks]: Using the $v-t$ graph for a body moving with uniform acceleration $a$, derive the equation $s = ut + \frac{1}{2}at^2$.

Velocity-Time Graph for Uniform Acceleration

Step 1: Identify Displacement as Area.
The total displacement ($s$) is given by the area under the velocity-time graph for the time interval $t$.

$$s = \text{Area of trapezium } OABD$$

$$s = \text{Area of rectangle } OACD + \text{Area of triangle } ABC$$

Step 2: Calculate Area of the Rectangle.
The rectangle has a length of $t$ (from $O$ to $D$) and a height of $u$ (from $O$ to $A$).

$$\text{Area of rectangle } = \text{length} \times \text{height} = t \times u = ut$$

Step 3: Calculate Area of the Triangle.
The triangle has a base of $t$ (length $AC$) and a height of $(v - u)$ (length $BC$).

$$\text{Area of triangle } = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times t \times (v - u)$$

Step 4: Use the Definition of Acceleration.
We know that acceleration $a$ is the change in velocity over time: $a = \frac{v - u}{t}$.
Therefore, we can substitute $(v - u) = at$.

$$\text{Area of triangle } = \frac{1}{2} \times t \times (at) = \frac{1}{2}at^2$$

Step 5: Combine the Areas.
Add the area of the rectangle and the area of the triangle together to get the total displacement.

$$s = ut + \frac{1}{2}at^2 \quad (\text{Proved})$$

Q4 [5 Marks]: A particle starts from the origin. Its acceleration is $a = 4 - 2t$. Draw an approximate $a-t$, $v-t$, and $x-t$ graph for the particle for the first $4\text{ s}$, assuming it starts from rest. At what time does it reverse its direction of motion?

Step 1: Analyze the Equations of Motion.

• Acceleration: $a(t) = 4 - 2t$. This is a straight line starting at $4$ and crossing zero at $t=2\text{ s}$.
• Velocity: $v(t) = \int a(t) dt = \int (4 - 2t) dt = 4t - t^2$ (since starts from rest, $v(0)=0$). This is a downward-opening parabola. Velocity is zero when $4t - t^2 = 0 \Rightarrow t(4 - t) = 0$, so at $t=0$ and $t=4\text{ s}$.
• Position: $x(t) = \int v(t) dt = \int (4t - t^2) dt = 2t^2 - \frac{t^3}{3}$ (since starts at origin, $x(0)=0$). This is a cubic curve.

Step 2: Reversal of Direction.
A particle reverses direction when its velocity crosses zero (changing from positive to negative, or vice versa). From our velocity equation $v(t) = t(4-t)$, velocity is positive from $t=0$ to $t=4$, and becomes negative right after $t=4$.

$$\text{The particle reverses direction at } t = 4\text{ s}$$

Acceleration ($a-t$)

Velocity ($v-t$)

Position ($x-t$)