1. **Stating the problem:** Calculate the deceleration of the car during the first 8 seconds and find the time $t$ when the total distance travelled by the car is 245 m. 2. **Find deceleration in the first 8 seconds:** - Initial speed $u = 15\,\text{ms}^{-1}$ - Speed at $8$ seconds $v = 10\,\text{ms}^{-1}$ - Time interval $\Delta t = 8\,\text{s}$ Using the formula for acceleration (or deceleration): $$a = \frac{v - u}{\Delta t} = \frac{10 - 15}{8} = \frac{-5}{8} = -0.625\,\text{ms}^{-2}$$ The negative sign indicates deceleration. 3. **Calculate distance travelled in first two intervals:** - From $0$ to $8$ seconds (speed reduces from 15 to 10): Using average speed: $$s_1 = \text{average speed} \times \Delta t = \frac{15 + 10}{2} \times 8 = 12.5 \times 8 = 100\,\text{m}$$ - From $8$ to $20$ seconds (speed constant at 10 ms$^{-1}$): $$s_2 = 10 \times (20 - 8) = 10 \times 12 = 120\,\text{m}$$ 4. **Calculate distance in last interval ($20$ to $t$ seconds):** - Speed decreases from 10 ms$^{-1}$ at $20$s to 0 ms$^{-1}$ at $t$s. - Time interval: $t - 20$ Average speed in this interval: $$\frac{10 + 0}{2} = 5\,\text{ms}^{-1}$$ Distance covered: $$s_3 = 5 (t - 20)\,\text{m}$$ 5. **Total distance travelled is 245 m:** $$s_1 + s_2 + s_3 = 245$$ Substitute values: $$100 + 120 + 5 (t - 20) = 245$$ Simplify: $$220 + 5t - 100 = 245$$ $$5t + 120 = 245$$ $$5t = 125$$ $$t = 25\,\text{s}$$ **Final answers:** (a) Deceleration $= -0.625\,\text{ms}^{-2}$ (b) Time $t = 25$ seconds