set 12k

Q1 [3 Marks]: What is the significance of the area enclosed by an acceleration-time ($a-t$) graph?

Answer:
The area enclosed by an acceleration-time ($a-t$) graph and the time axis represents the change in velocity ($\Delta v$) of the object during that specific time interval.

Mathematical Proof:
By definition, acceleration is the rate of change of velocity:

$$a = \frac{dv}{dt}$$

Separating the variables gives:

$$dv = a \, dt$$

Integrating both sides from initial time $t_1$ to final time $t_2$:

$$\int_{v_1}^{v_2} dv = \int_{t_1}^{t_2} a \, dt$$

$$v_2 - v_1 = \int_{t_1}^{t_2} a \, dt$$

The integral on the right is exactly the definition of the area under the $a-t$ curve. Therefore, Area = $\Delta v$.

Q2 [3 Marks]: Is it possible for the distance-time graph of a particle to have a negative slope? Explain why or why not.

Answer: NO.

Explanation:
Distance is a scalar quantity representing the total path length covered by a particle. Unlike displacement, distance can never decrease with time. It can only increase (if the particle is moving) or remain constant (if the particle is at rest).

The slope of a distance-time graph represents the instantaneous speed ($\frac{ds}{dt}$) of the particle. Because distance never decreases, the slope can never be negative. (Note: A displacement-time graph can have a negative slope, which represents a negative velocity).

Q3 [4 Marks]: A body covers $12\text{ m}$ in the 2nd second and $20\text{ m}$ in the 4th second. How much distance will it cover in 4 seconds after the 5th second?

Step 1: Use the $n^{th}$ second formula.
The distance covered in the $n^{th}$ second is given by: $S_n = u + \frac{a}{2}(2n - 1)$.

Step 2: Formulate equations for the 2nd and 4th seconds.
For the 2nd second ($n=2$):

$$12 = u + \frac{a}{2}(2(2) - 1) \Rightarrow u + 1.5a = 12 \quad \text{--- (Eq 1)}$$

For the 4th second ($n=4$):

$$20 = u + \frac{a}{2}(2(4) - 1) \Rightarrow u + 3.5a = 20 \quad \text{--- (Eq 2)}$$

Step 3: Solve for $a$ and $u$.
Subtract Equation 1 from Equation 2:

$$(u + 3.5a) - (u + 1.5a) = 20 - 12$$

$$2a = 8 \Rightarrow a = 4\text{ m/s}^2$$

Substitute $a = 4$ into Equation 1:

$$u + 1.5(4) = 12 \Rightarrow u + 6 = 12 \Rightarrow u = 6\text{ m/s}$$

Step 4: Analyze the required interval.
We need the distance covered in the 4 seconds after the 5th second (i.e., from $t=5\text{ s}$ to $t=9\text{ s}$).
First, find the velocity of the body exactly at $t=5\text{ s}$. This becomes our new initial velocity ($u'$) for this interval.

$$v_5 = u + at = 6 + 4(5) = 26\text{ m/s}$$

Step 5: Calculate the distance.
Use $S = u't + \frac{1}{2}at^2$ for the 4-second interval starting at $t=5$:

$$S = 26(4) + \frac{1}{2}(4)(4)^2$$

$$S = 104 + 2(16) = 104 + 32$$

$$\text{Distance } = 136\text{ m}$$

Q4 [5 Marks]: The velocity-time graph of a particle moving in a straight line consists of a triangle with a peak velocity of $10\text{ m/s}$ at $t=2\text{ s}$, coming back to zero at $t=4\text{ s}$. Calculate the distance traveled in the first $2\text{ s}$ and the acceleration during the interval $t=2$ to $t=4\text{ s}$.

Velocity-Time Graph (Triangle)

Part 1: Distance traveled in the first $2\text{ s}$
The distance is the area under the $v-t$ graph from $t=0$ to $t=2\text{ s}$. This area is a right-angled triangle.

$$\text{Distance } = \frac{1}{2} \times \text{base} \times \text{height}$$

$$\text{Distance } = \frac{1}{2} \times (2 - 0) \times 10$$

$$\text{Distance } = \frac{1}{2} \times 2 \times 10$$

$$\text{Distance in first 2s } = 10\text{ m}$$

Part 2: Acceleration during $t=2$ to $t=4\text{ s}$
Acceleration is the slope of the $v-t$ graph. During this interval, the velocity goes from $v_1 = 10\text{ m/s}$ (at $t_1 = 2$) down to $v_2 = 0\text{ m/s}$ (at $t_2 = 4$).

$$a = \frac{v_2 - v_1}{t_2 - t_1}$$

$$a = \frac{0 - 10}{4 - 2} = \frac{-10}{2}$$

$$\text{Acceleration } = -5\text{ m/s}^2$$

(The negative sign correctly indicates deceleration/retardation as the slope is downward).