A particle starts from rest and moves with a constant acceleration of 0.5ms⁻². Calculate the time taken by the body to cover a distance of 25m (SSCE). A car slows down from 108 kmh⁻¹ with a uniform retardation of 5ms⁻² (i) How long will it take to reach 18kmh⁻¹ (ii) What is the distance covered?

Here are the solutions to the problems:

:
Given:
Initial velocity, $u = 0 \, \text{ms}^{-1}$ (starts from rest)
Acceleration, $a = 0.5 \, \text{ms}^{-2}$
Distance, $s = 25 \, \text{m}$

Step 1: Identify the appropriate kinematic equation.
Since we have $u$, $a$, $s$ and need to find $t$, we use the equation:

$$s = ut + \frac{1}{2}at^2$$

Step 2: Substitute the given values into the equation.

$$25 = (0)t + \frac{1}{2}(0.5)t^2$$ $$25 = 0.25t^2$$

Step 3: Solve for $t$.

$$t^2 = \frac{25}{0.25}$$ $$t^2 = 100$$ $$t = \sqrt{100}$$ $$t = 10.0 \, \text{s}$$

The time taken is $\boxed{\text{10.0 s}}$.

:
Given:
Initial velocity, $u = 108 \, \text{kmh}^{-1}$
Final velocity, $v = 18 \, \text{kmh}^{-1}$
Retardation (negative acceleration), $a = -5 \, \text{ms}^{-2}$

Step 1: Convert velocities from $\text{kmh}^{-1}$ to $\text{ms}^{-1}$.
To convert $\text{kmh}^{-1}$ to $\text{ms}^{-1}$, multiply by $\frac{1000}{3600}$ or $\frac{5}{18}$.

$$u = 108 \, \text{kmh}^{-1} = 108 \times \frac{5}{18} \, \text{ms}^{-1} = 6 \times 5 \, \text{ms}^{-1} = 30 \, \text{ms}^{-1}$$ $$v = 18 \, \text{kmh}^{-1} = 18 \times \frac{5}{18} \, \text{ms}^{-1} = 5 \, \text{ms}^{-1}$$