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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.

Ask by Bates Gibbs. in Ghana
Mar 14,2025

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Answer

(a) Distance traveled: 275 meters (b) Average speed: approximately 7.86 m/s (c) Time to cover half the distance: 18.75 seconds

Solution

To solve the problem, we will break it down into three parts based on the car's motion: acceleration, uniform speed, and deceleration. ### Known Conditions: 1. **Acceleration phase:** - Initial speed \( u = 0 \, \text{m/s} \) - Acceleration \( a = 1 \, \text{m/s}^2 \) - Time \( t_1 = 10 \, \text{s} \) 2. **Uniform speed phase:** - Time \( t_2 = 20 \, \text{s} \) 3. **Deceleration phase:** - Time \( t_3 = 5 \, \text{s} \) - Final speed \( v = 0 \, \text{m/s} \) ### Step-by-Step Solution: #### (a) Distance travelled 1. **Distance during acceleration:** The distance \( s_1 \) during the acceleration phase can be calculated using the formula: \[ s_1 = ut + \frac{1}{2} a t^2 \] Substituting the known values: \[ s_1 = 0 \cdot 10 + \frac{1}{2} \cdot 1 \cdot (10)^2 = \frac{1}{2} \cdot 1 \cdot 100 = 50 \, \text{m} \] 2. **Distance during uniform speed:** The speed at the end of the acceleration phase is given by: \[ v = u + at = 0 + 1 \cdot 10 = 10 \, \text{m/s} \] The distance \( s_2 \) during the uniform speed phase is: \[ s_2 = v \cdot t_2 = 10 \cdot 20 = 200 \, \text{m} \] 3. **Distance during deceleration:** The initial speed for the deceleration phase is \( v = 10 \, \text{m/s} \) and the final speed is \( 0 \, \text{m/s} \). The average speed during deceleration is: \[ v_{\text{avg}} = \frac{v + 0}{2} = \frac{10 + 0}{2} = 5 \, \text{m/s} \] The distance \( s_3 \) during the deceleration phase is: \[ s_3 = v_{\text{avg}} \cdot t_3 = 5 \cdot 5 = 25 \, \text{m} \] 4. **Total distance travelled:** \[ s_{\text{total}} = s_1 + s_2 + s_3 = 50 + 200 + 25 = 275 \, \text{m} \] #### (b) Average speed The total time taken \( T \) is: \[ T = t_1 + t_2 + t_3 = 10 + 20 + 5 = 35 \, \text{s} \] The average speed \( v_{\text{avg}} \) is given by: \[ v_{\text{avg}} = \frac{s_{\text{total}}}{T} = \frac{275}{35} \approx 7.857 \, \text{m/s} \] #### (c) Time taken to cover half the distance Half the total distance is: \[ \frac{s_{\text{total}}}{2} = \frac{275}{2} = 137.5 \, \text{m} \] 1. **Distance covered in the acceleration phase:** - The car covers \( 50 \, \text{m} \) in the first \( 10 \, \text{s} \). 2. **Distance covered in the uniform speed phase:** - The car covers \( 200 \, \text{m} \) in the next \( 20 \, \text{s} \), reaching a total of \( 250 \, \text{m} \) by the end of this phase. Since \( 137.5 \, \text{m} \) is less than \( 250 \, \text{m} \), we need to find the time taken to cover the remaining distance after the acceleration phase. 3. **Remaining distance after acceleration:** \[ \text{Remaining distance} = 137.5 - 50 = 87.5 \, \text{m} \] 4. **Time taken to cover \( 87.5 \, \text{m} \) at \( 10 \, \text{m/s} \):** \[ t = \frac{\text{distance}}{\text{speed}} = \frac{87.5}{10} = 8.75 \, \text{s} \] 5. **Total time to cover half the distance:** \[ \text{Total time} = t_1 + t = 10 + 8.75 = 18.75 \, \text{s} \] ### Final Answers: (a) Distance travelled: \( 275 \, \text{m} \) (b) Average speed: \( \approx 7.857 \, \text{m/s} \) (c) Time taken to cover half the distance: \( 18.75 \, \text{s} \)

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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.**

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