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प्रश्न
Complete the following table:
| Car Model | Driver X Reaction time 0.20 s |
Driver Y Reaction time 0.30 s |
| A (deceleration on hard braking = 6.0 m/s2) | Speed = 54 km/h Braking distance a = ............ Total stopping distance b = ............ |
Speed = 72 km/h Braking distance c = ........... Total stopping distance d = ............ |
| B (deceleration on hard braking = 7.5 m/s2) | Speed = 54 km/h Breaking distance e = ........... Total stopping distance f = ............ |
Speed 72 km/h Braking distance g = ............. Total stopping distance h = ............ |
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उत्तर
Braking distance: Distance travelled after the brakes are applied.
Total stopping distance = Braking distance + Distance travelled in the reaction time
Case A:
Deceleration = 6.0 m/s2
For driver X:
Initial velocity, u = 54 km/h = 15 m/s
Final velocity, v = 0
Braking distance,
Distance travelled in the reaction time = 15 × 0.20 = 3 m
Total stopping distance, b = 19 + 3 = 22 m
For driver Y:
Initial velocity, u = 72 km/h = 20 m/s
Final velocity, v = 0
Braking distance,
Total stopping distance, d = 33 + 6 = 39 m
Case B:
Deceleration = 6.0 m/s2
Now, we have:
e = 15 m
f = 18 m
g = 27 m
h = 33 m
| Car Model | Driver X Reaction Time = 0.20 s |
Driver Y Reaction Time = 0.30 s |
| A (deceleration on hard braking = 6.0 m/s2) | Speed = 54 km/h Braking distance, a = 19 m Total stopping distance, b = 22 m |
Speed = 72 km/h Braking distance, c = 33 m Total stopping distance, d = 39 m |
| B (deceleration on hard braking = 7.5 m/s2) | Speed = 54 km/h Breaking distance, e = 15 m Total stopping distance, f = 18 m |
Speed = 72 km/h Braking distance, g = 27 m Total stopping distance, h = 33 m |
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