A car starts from rest and its velocity reaches 24 ms⁻¹ in 6 s. Find the average acceleration and the distance travelled in these 6 s.

Given:

Initial velocity, $u = 0 \text{ ms}^{-1}$

Final velocity, $v = 24 \text{ ms}^{-1}$

Time taken, $t = 6 \text{ s}$

Average acceleration can be calculated by using the formula:

$$a = \frac{v - u}{t}$$

$$a = \frac{24 - 0}{6}$$

$$a = 4 \text{ ms}^{-2}$$

Therefore, the average acceleration is $4 \text{ ms}^{-2}$. Distance travelled can be calculated by assuming uniform acceleration as:

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

$$s = (0 \times 6) + \frac{1}{2} \times 4 \times 6^2$$

$$s = 36 \times 2$$

$$s = 72 \text{ m}$$

Therefore, the distance travelled in 6 s is 72 m.