14. A car starts from rest and accelerates at \( 1 \mathrm{~ms}^{-2} \) for 10 seconds. It then continues at a uniform speed for another 20 seconds and decelerates to rest in 5 seconds. Find the: (a) distance travelled; (b) average speed; (c) time taken to cover half the distance.

Let's break this down step by step! First, during the acceleration phase, the car accelerates at \(1 \, \text{m/s}^2\) for 10 seconds. Using the formula for distance during uniform acceleration: \[ d_1 = \frac{1}{2} a t^2 = \frac{1}{2} \times 1 \times 10^2 = \frac{1}{2} \times 1 \times 100 = 50 \, \text{m} \] Next, the speed of the car at the end of this phase is: \[ v = at = 1 \times 10 = 10 \, \text{m/s} \] In the next phase, for 20 seconds at a uniform speed, the distance covered is: \[ d_2 = v \times t = 10 \times 20 = 200 \, \text{m} \] Finally, during the deceleration phase, it takes 5 seconds to come to a stop from 10 m/s. The distance covered while decelerating can be calculated as: \[ d_3 = v \times t - \frac{1}{2} a t^2 = 10 \times 5 - \frac{1}{2} \times 2 \times 5^2 = 50 - 25 = 25 \, \text{m} \] Now, summing up the distances: \[ \text{Total Distance} = d_1 + d_2 + d_3 = 50 + 200 + 25 = 275 \, \text{m} \] **(a) Distance travelled = 275 m.** For the average speed: The total time taken is: \[ t_{\text{total}} = 10 + 20 + 5 = 35 \, \text{s} \] Average speed can be calculated using the formula: \[ \text{Average speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{275}{35} \approx 7.857 \, \text{m/s} \] **(b) Average speed = 7.857 m/s.** To find the time taken to cover half the distance, we first note that half the distance is: \[ \frac{275}{2} = 137.5 \, \text{m} \] During the acceleration phase, the car travels 50 m in 10 seconds, and during the constant speed phase, the car covers 200 m in 20 seconds. After 10 seconds, the car has covered 50 m, so: \[ 137.5 - 50 = 87.5 \, \text{m} \] At 10 m/s during the constant speed phase, the time required to travel the remaining 87.5 m is: \[ t = \frac{\text{Distance}}{\text{Speed}} = \frac{87.5}{10} = 8.75 \, \text{s} \] Therefore, the total time to cover half the distance is: \[ 10 \, \text{s} + 8.75 \, \text{s} = 18.75 \, \text{s} \] **(c) Time taken to cover half the distance = 18.75 s.**