A train starting from rest picks up a speed of 20 m/s in 20 s while travelling on a straight path. It continues to move at the same speed for the next 50 s. It is then brought to rest in the next 10 s. a. Plot the speed-time graph for the train. b. Calculate its acceleration and retardation. c. Calculate the distance covered by the train when it was i) travelling at the same speed, ii) retarding. d. Calculate the average speed during retardation. e. How much distance did the train cover during this entire journey?
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Answer:
Explanation:
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Verified answer
Answer:
a. To plot the speed-time graph for the train, we can break down the given information into three phases: acceleration, constant speed, and deceleration.
1. Acceleration phase (0 to 20 s):
- Initial speed (u) = 0 m/s
- Final speed (v) = 20 m/s
- Time (t) = 20 s
2. Constant speed phase (20 to 70 s):
- Speed (v) = 20 m/s
- Time (t) = 50 s
3. Deceleration phase (70 to 80 s):
- Initial speed (u) = 20 m/s
- Final speed (v) = 0 m/s
- Time (t) = 10 s
Based on these phases, the speed-time graph can be plotted as follows:
```
^
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20 | .
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+----------------------------------->
0 20 70 80 Time (s)
```
b. Acceleration and retardation:
Acceleration (a) during the acceleration phase:
a = (v - u) / t
= (20 m/s - 0 m/s) / 20 s
= 1 m/s²
Retardation (negative acceleration) during the deceleration phase:
a = (v - u) / t
= (0 m/s - 20 m/s) / 10 s
= -2 m/s²
c. Distance covered:
i) Distance covered when traveling at the same speed:
Distance = Speed × Time
Distance = 20 m/s × 50 s
Distance = 1000 m
ii) Distance covered during deceleration:
Distance = (Initial speed × Time) + (0.5 × Acceleration × Time²)
Distance = (20 m/s × 10 s) + (0.5 × (-2 m/s²) × (10 s)²
Distance = 200 m - 100 m
Distance = 100 m
d. Average speed during deceleration:
Average speed = (Initial speed + Final speed) / 2
Average speed = (20 m/s + 0 m/s) / 2
Average speed = 10 m/s
e. Total distance covered during the entire journey:
Distance during acceleration = (Initial speed × Time) + (0.5 × Acceleration × Time²)
Distance during acceleration = (0 m/s × 20 s) + (0.5 × 1 m/s² × (20 s)²)
Distance during acceleration = 200 m
Distance during constant speed = Speed × Time
Distance during constant speed = 20 m/s × 50 s
Distance during constant speed = 1000 m
Distance during deceleration = (Initial speed × Time) + (0.5 × Retardation × Time²)
Distance during deceleration = (20 m/s × 10 s) + (0.5 × (-2 m/s²) × (10 s)²)
Distance during deceleration = 200 m - 100 m
Distance during deceleration = 100 m
Total distance covered = Distance during acceleration + Distance during constant speed + Distance during deceleration
Total distance covered =
200 m + 1000 m + 100 m
Total distance covered = 1300 m
Therefore, the train covered a total distance of 1300 meters during its entire journey.