set 14k

Q1 [3 Marks]: If the initial velocity of a vehicle is tripled, by what factor does its stopping distance increase, assuming the braking retardation remains constant?

Step 1: Write the stopping distance formula.
When a vehicle comes to a stop, its final velocity $v = 0$. Using $v^2 = u^2 - 2as$ (where $a$ is the magnitude of retardation):

$$s = \frac{u^2}{2a}$$

Step 2: Apply the new condition.
The new initial velocity is $u' = 3u$. The retardation $a$ remains the same.

$$s' = \frac{(u')^2}{2a} = \frac{(3u)^2}{2a}$$

$$s' = \frac{9u^2}{2a}$$

Step 3: Compare the new distance to the original distance.

$$s' = 9 \left( \frac{u^2}{2a} \right) = 9s$$

$$\text{The stopping distance increases by a factor of 9.}$$

Q2 [3 Marks]: A bullet loses $1/n$ of its velocity passing through one wooden plank. What is the minimum number of identical planks required to completely stop the bullet?

Step 1: Analyze motion through the first plank.
Let initial velocity be $u$. It loses $\frac{1}{n}$ of its velocity, so it retains $\left(1 - \frac{1}{n}\right)$ of it.
Final velocity after one plank: $v_1 = u - \frac{u}{n} = u\left(\frac{n-1}{n}\right)$.
Let the thickness of one plank be $x$ and the constant retardation be $a$. Using $v^2 = u^2 - 2ax$:

$$\left[ u\left(\frac{n-1}{n}\right) \right]^2 = u^2 - 2ax$$

$$u^2 \left( \frac{n^2 - 2n + 1}{n^2} \right) = u^2 - 2ax$$

Step 2: Isolate the retardation term ($2ax$).

$$2ax = u^2 - u^2 \left( \frac{n^2 - 2n + 1}{n^2} \right)$$

$$2ax = u^2 \left[ 1 - \left( \frac{n^2 - 2n + 1}{n^2} \right) \right]$$

$$2ax = u^2 \left[ \frac{n^2 - (n^2 - 2n + 1)}{n^2} \right] = u^2 \left( \frac{2n - 1}{n^2} \right)$$

Step 3: Analyze motion to completely stop the bullet.
To completely stop the bullet, the final velocity must be $0$. Let $N$ be the total number of planks needed. The total distance is $Nx$.

$$0 = u^2 - 2a(Nx)$$

$$N(2ax) = u^2 \Rightarrow N = \frac{u^2}{2ax}$$

Step 4: Substitute $2ax$ to find $N$.

$$N = \frac{u^2}{u^2 \left( \frac{2n - 1}{n^2} \right)} = \frac{n^2}{2n - 1}$$

$$\text{Number of Planks } (N) = \frac{n^2}{2n - 1}$$

Q3 [4 Marks]: An object covers $9\text{ m}$ in the first $3\text{ s}$ and $24\text{ m}$ in the next $3\text{ s}$. Find its initial velocity and uniform acceleration.

Step 1: Write equation for the first interval.
Distance $s_1 = 9\text{ m}$, time $t_1 = 3\text{ s}$. Use $s = ut + \frac{1}{2}at^2$.

$$9 = u(3) + \frac{1}{2}a(3)^2$$

$$9 = 3u + 4.5a$$

Divide the entire equation by 3 to simplify:

$$u + 1.5a = 3 \quad \text{--- (Equation 1)}$$

Step 2: Write equation for the total journey (both intervals).
It covers $24\text{ m}$ in the next $3\text{ s}$. This means in a total time of $t_2 = 3 + 3 = 6\text{ s}$, it covers a total distance of $s_2 = 9 + 24 = 33\text{ m}$.

$$33 = u(6) + \frac{1}{2}a(6)^2$$

$$33 = 6u + 18a$$

Divide the entire equation by 6 to simplify:

$$u + 3a = 5.5 \quad \text{--- (Equation 2)}$$

Step 3: Solve the simultaneous equations.
Subtract Equation 1 from Equation 2:

$$(u + 3a) - (u + 1.5a) = 5.5 - 3$$

$$1.5a = 2.5 \Rightarrow a = \frac{2.5}{1.5} = \frac{5}{3}$$

$$\text{Acceleration } (a) = 1.67\text{ m/s}^2$$

Step 4: Find initial velocity ($u$).
Substitute $a = \frac{5}{3}$ back into Equation 1:

$$u + 1.5\left(\frac{5}{3}\right) = 3$$

$$u + \frac{3}{2}\left(\frac{5}{3}\right) = 3 \Rightarrow u + 2.5 = 3$$

$$\text{Initial Velocity } (u) = 0.5\text{ m/s}$$

Q4 [5 Marks]: A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of $3.0\text{ m/s}^2$ for $8.0\text{ s}$. How far does the boat travel during this time? If it then runs out of fuel and experiences a water drag deceleration of $0.5\text{ m/s}^2$, what is the total distance traveled from the start?

Part 1: Powered Phase (Engine On)
Initial velocity $u = 0$, acceleration $a_1 = 3.0\text{ m/s}^2$, time $t_1 = 8.0\text{ s}$.
Distance traveled ($s_1$):

$$s_1 = ut_1 + \frac{1}{2}a_1 t_1^2 = 0 + \frac{1}{2}(3.0)(8.0)^2$$

$$s_1 = 1.5 \times 64 = 96\text{ m}$$

Find the velocity at the end of this phase ($v_1$), which will be the initial velocity for the next phase.

$$v_1 = u + a_1 t_1 = 0 + 3.0(8.0) = 24\text{ m/s}$$

Velocity-Time Graph (Motorboat)

Part 2: Coasting Phase (Water Drag)
The boat is moving at $u_2 = 24\text{ m/s}$. The engine is off, so acceleration is $a_2 = -0.5\text{ m/s}^2$ until it stops ($v_2 = 0$).
Using the formula $v^2 = u^2 + 2as$:

$$0 = (24)^2 + 2(-0.5)s_2$$

$$0 = 576 - 1.0(s_2) \Rightarrow s_2 = 576\text{ m}$$

(Notice on the graph how the shallow drag slope creates a massive stopping distance!)

Part 3: Total Distance

$$D_{total} = s_1 + s_2 = 96 + 576$$

$$\text{Total Distance Traveled } = 672\text{ m}$$