LP Grade 10 Physical Sciences...

Here are the solutions to Question 4.

4.1 Explain the term 'slows down uniformly at $2.5 m \cdot s^{-2}$'.

This means the car is undergoing a constant deceleration of $2.5 m \cdot s^{-2}$. Its velocity decreases by $2.5 \text{ m/s}$ every second.

4.2 Show that $72 km \cdot h^{-1} = 20 m \cdot s^{-1}$.

Step 1: Convert kilometers to meters and hours to seconds. $1 km = 1000 m$ $1 h = 3600 s$

Step 2: Apply the conversion factors to the given speed. $72 km \cdot h^{-1} = 72 \times \frac{1000 m}{1 km} \times \frac{1 h}{3600 s}$ $72 km \cdot h^{-1} = \frac{72 \times 1000}{3600} m \cdot s^{-1}$ $72 km \cdot h^{-1} = \frac{72000}{3600} m \cdot s^{-1}$ $72 km \cdot h^{-1} = 20 m \cdot s^{-1}$ This shows that $\boxed{72 km \cdot h^{-1} = 20 m \cdot s^{-1}}$.

4.3 Determine the distance travelled by the motor car during the $0.4 \text{ s}$ reaction time.

During the reaction time, the car travels at a constant speed.

Step 1: Identify the constant speed and reaction time. Initial speed ($v$) = $20 m \cdot s^{-1}$ (from 4.2) Reaction time ($\Delta t$) = $0.4 \text{ s}$

Step 2: Calculate the distance using the formula: distance = speed $\times$ time. $Distance_{reaction} = v \times \Delta t$ $Distance_{reaction} = 20 m \cdot s^{-1} \times 0.4 s$ $Distance_{reaction} = 8 m$ The distance travelled during the reaction time is $\boxed{8 m}$.

4.4 Calculate the total time taken for the motor car to come to rest, from the instant that the driver saw the landmine exploding.

The total time consists of the reaction time and the braking time.

Step 1: Determine the time taken for the car to come to rest after applying brakes (braking time). Initial velocity ($u$) = $20 m \cdot s^{-1}$ Final velocity ($v$) = $0 m \cdot s^{-1}$ (car comes to rest) Acceleration ($a$) = $-2.5 m \cdot s^{-2}$ (deceleration) Using the equation of motion: $v = u + a \Delta t_{braking}$ $0 = 20 m \cdot s^{-1} + (-2.5 m \cdot s^{-2}) \Delta t_{braking}$ $2.5 \Delta t_{braking} = 20$ $\Delta t_{braking} = \frac{20}{2.5} = 8 s$

Step 2: Calculate the total time. $Total time = Reaction time + Braking time$ $Total time = 0.4 s + 8 s$ $Total time = 8.4 s$ The total time taken for the motor car to come to rest is $\boxed{8.4 s}$.

4.5 Will the motor car stop before reaching the wreck of the truck? Show all working in your answer.

To answer this, we need to calculate the total stopping distance of the car and compare it to the initial distance between the car and the truck.

Step 1: Calculate the distance travelled during braking. Initial velocity ($u$) = $20 m \cdot s^{-1}$ Final velocity ($v$) = $0 m \cdot s^{-1}$ Acceleration ($a$) = $-2.5 m \cdot s^{-2}$ Using the equation of motion: $v^2 = u^2 + 2a \Delta x_{braking}$ $(0 m \cdot s^{-1})^2 = (20 m \cdot s^{-1})^2 + 2(-2.5 m \cdot s^{-2}) \Delta x_{braking}$ $0 = 400 - 5 \Delta x_{braking}$ $5 \Delta x_{braking} = 400$ $\Delta x_{braking} = \frac{400}{5} = 80 m$

Step 2: Calculate the total stopping distance. $Total stopping distance = Distance_{reaction} + Distance_{braking}$ $Total stopping distance = 8 m + 80 m$ $Total stopping distance = 88 m$

Step 3: Compare the total stopping distance to the initial distance between the car and the truck. The initial distance between the car and the truck is $100 \text{ m}$. Since $88 m < 100 \text{ m}$, the motor car will stop before reaching the wreck of the truck. The motor car will stop $\boxed{before reaching the wreck of the truck}$.