Set 6 k

Q1 [3 Marks]: Explain the concept of stopping distance for a moving vehicle. Write the formula relating stopping distance to initial velocity and retardation.

Part A: Concept Explanation
Stopping distance is the total distance traveled by a moving vehicle from the moment the driver decides to stop until the vehicle completely comes to rest. It consists of two parts:

1. Reaction Distance: The distance the vehicle travels uniformly during the driver's reaction time (before brakes are physically applied).
2. Braking Distance: The distance the vehicle travels while decelerating under the action of the brakes.

The formula typically refers to just the braking distance component unless reaction time is specified.

Part B: Formula
For the braking distance portion: Let initial velocity be $u$, final velocity $v=0$, and retardation be $a$. Using the third equation of motion $v^2 = u^2 + 2as$:

$$0 = u^2 - 2ad \quad \text{(where a is magnitude of retardation)}$$

$$\text{Stopping Distance } (d) = \frac{u^2}{2a}$$

Q2 [3 Marks]: Water drops fall at regular intervals from a tap $5\text{ m}$ above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant?

Step 1: Find the time taken by the first drop to reach the ground.
Distance $s = 5\text{ m}$, initial velocity $u = 0$, $a = g = 10\text{ m/s}^2$ (downward is positive).

$$s = ut + \frac{1}{2}gt^2 \Rightarrow 5 = 0 + \frac{1}{2}(10)t^2$$

$$5 = 5t^2 \Rightarrow t^2 = 1 \Rightarrow t = 1\text{ s}$$

So, the total time for one drop to fall $5\text{ m}$ is $1\text{ s}$.

Step 2: Determine the time interval between drops.
The 3rd drop is leaving at $t=1\text{ s}$. This means there have been 2 intervals since the 1st drop fell (Drop 1 to Drop 2, then Drop 2 to Drop 3).
Time interval between drops = $\frac{1\text{ s}}{2} = 0.5\text{ s}$.

Step 3: Calculate the distance fallen by the second drop.
At $t=1\text{ s}$ (when Drop 1 hits ground), Drop 2 has been falling for exactly one interval, which is $0.5\text{ s}$.

$$s_2 = ut + \frac{1}{2}gt^2 = 0 + \frac{1}{2}(10)(0.5)^2$$

$$s_2 = 5(0.25) = 1.25\text{ m}$$

This is the distance it has fallen from the tap.

Step 4: Find the height above the ground.
Height = Total height - Distance fallen.

$$\text{Height} = 5\text{ m} - 1.25\text{ m}$$

$$\text{Height above ground } = 3.75\text{ m}$$

Q3 [4 Marks]: A ball is thrown vertically upwards. Draw its velocity-time graph for the complete journey (upward and downward). Indicate the point of maximum height on the graph.

Analysis:
Let upward be positive (+). The ball starts with a high positive velocity ($+u$). Gravity acts downwards ($a = -g$) as a constant acceleration. Therefore, the velocity decreases linearly over time until it becomes zero at maximum height. After that, it falls back down, so velocity becomes negative and increases in magnitude.
This results in a straight line with a constant negative slope intersecting the time axis.

Velocity-Time Graph (Vertical Throw)

Key Features:

• The straight line indicates constant acceleration (gravity).
• The y-intercept represents the initial launch velocity ($+u$).
• The x-intercept represents the point of maximum height where velocity is instantaneously zero.

Q4 [5 Marks]: A rocket is fired vertically upwards with a net acceleration of $4\text{ m/s}^2$. After $10\text{ s}$, its fuel is exhausted and it continues to move as a free-falling body. Calculate the maximum height reached and the total time taken to return to the ground. ($g=9.8\text{ m/s}^2$)

Part 1: Powered Ascent (First $10\text{ s}$)
Initial velocity $u = 0$, acceleration $a = 4\text{ m/s}^2$, time $t_1 = 10\text{ s}$.
Find velocity at fuel exhaustion ($v_1$):

$$v_1 = u + at_1 = 0 + 4(10) = 40\text{ m/s}$$

Find height reached during powered flight ($h_1$):

$$h_1 = ut_1 + \frac{1}{2}at_1^2 = 0 + \frac{1}{2}(4)(10)^2 = 2(100) = 200\text{ m}$$

Part 2: Free Flight (Ascent under gravity)
The rocket is now at $200\text{ m}$ traveling upwards at $40\text{ m/s}$. The engine is off, so acceleration is now $a = -g = -9.8\text{ m/s}^2$. It will continue to rise until velocity becomes zero ($v_2 = 0$).
Find distance risen during free flight ($h_2$):

$$v_2^2 = v_1^2 + 2(-g)h_2 \Rightarrow 0 = (40)^2 - 2(9.8)h_2$$

$$19.6h_2 = 1600 \Rightarrow h_2 = \frac{1600}{19.6} \approx 81.63\text{ m}$$

Find time taken for this free ascent ($t_2$):

$$v_2 = v_1 - gt_2 \Rightarrow 0 = 40 - 9.8t_2 \Rightarrow t_2 = \frac{40}{9.8} \approx 4.08\text{ s}$$

Part 3: Maximum Height

$$H_{max} = h_1 + h_2 = 200 + 81.63$$

$$\text{Maximum Height } = 281.63\text{ m}$$

Part 4: Return Journey (Descent)
The rocket now falls from $H_{max} = 281.63\text{ m}$ to the ground. Initial velocity for descent $u = 0$, $a = 9.8\text{ m/s}^2$.

$$H_{max} = \frac{1}{2}gt_3^2 \Rightarrow 281.63 = \frac{1}{2}(9.8)t_3^2$$

$$4.9t_3^2 = 281.63 \Rightarrow t_3^2 = 57.47 \Rightarrow t_3 \approx 7.58\text{ s}$$

Part 5: Total Time of Flight

$$T_{total} = t_1 \text (powered) + t_2 \text (free ascent) + t_3 \text (descent)$$

$$T_{total} = 10 + 4.08 + 7.58$$

$$\text{Total Time } = 21.66\text{ s}$$