Set 10k

Q1 [3 Marks]: Define average and instantaneous acceleration. How can instantaneous acceleration be obtained from a velocity-time graph?

Part A: Definitions

• Average Acceleration ($\bar{a}$): It is defined as the total change in velocity divided by the total time interval over which that change occurred. Formula: $\bar{a} = \frac{\Delta v}{\Delta t}$.
• Instantaneous Acceleration ($a$): It is the exact acceleration of an object at a specific instant in time. Mathematically, it is the limit of average acceleration as the time interval approaches zero. Formula: $a = \frac{dv}{dt}$.

Part B: From a Graph
Instantaneous acceleration can be obtained from a velocity-time ($v-t$) graph by drawing a tangent line to the curve at that specific instant of time. The slope (gradient) of that tangent line represents the instantaneous acceleration.

Q2 [3 Marks]: The velocity-time graph of a moving object is a straight line inclined to the time axis. What does the slope and the area under this graph represent?

Analysis of the Graph:
A straight line inclined to the time axis on a $v-t$ graph implies that velocity is changing at a perfectly constant rate.

1. The Slope:
The slope of a velocity-time graph is $\frac{\Delta v}{\Delta t}$. Therefore, the slope represents the uniform (constant) acceleration of the object.

2. The Area:
The area enclosed between the velocity-time line and the time axis (x-axis) represents the total displacement of the object during that specific time interval. (If the line does not cross the time axis, it also represents the total distance traveled).

Q3 [4 Marks]: Draw the position-time graph for: (i) an object at rest, (ii) uniform motion, and (iii) uniform accelerated motion.

(i) Object at Rest

(ii) Uniform Motion

(iii) Uniform Acceleration

Q4 [5 Marks]: The velocity-time graph of a particle starting from rest reveals it accelerates uniformly at $2\text{ m/s}^2$ for $5\text{ s}$, then moves with uniform velocity for $10\text{ s}$, and finally retards uniformly to stop in $5\text{ s}$. Calculate the total distance covered and the average velocity of the particle.

Method: Using Graph Areas (Fastest Method)
First, let's find the maximum velocity the particle reaches after the initial $5\text{ s}$ of acceleration.

$$v_{max} = u + at_1 = 0 + (2)(5) = 10\text{ m/s}$$

The particle maintains this $10\text{ m/s}$ for $10\text{ s}$, and then decelerates back to $0$ in $5\text{ s}$.

Velocity-Time Graph (Trapezium)

Step 1: Calculate Total Distance.
The total distance is the area under the v-t graph. We can break it down into three geometric shapes:

• Area 1 (Acceleration Triangle): $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 10 = 25\text{ m}$
• Area 2 (Uniform Velocity Rectangle): $\text{width} \times \text{height} = 10 \times 10 = 100\text{ m}$
• Area 3 (Retardation Triangle): $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 10 = 25\text{ m}$

$$D_{total} = 25 + 100 + 25 = 150\text{ m}$$

(Alternatively, Area of Trapezium = $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} = \frac{1}{2} \times (10 + 20) \times 10 = 150\text{ m}$).

$$\text{Total Distance } = 150\text{ m}$$

Step 2: Calculate Average Velocity.
Total time taken for the entire journey $t_{total} = 5 + 10 + 5 = 20\text{ s}$.

$$v_{avg} = \frac{\text{Total Displacement}}{\text{Total Time}}$$

$$v_{avg} = \frac{150\text{ m}}{20\text{ s}}$$

$$\text{Average Velocity } = 7.5\text{ m/s}$$