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Can you explain the answer of this question below:
A man is moving with 36 kmph. The time of reaction is 0.9 seconds. On seeing an obstacle in the path, he applies brakes and decelerates at 5 m/s2, the total distance covered before he stops is:
- A:19 m
- B:17 m
- C:16 m
- D:18 m
The answer is a.
Most Upvoted Answer
Can you explain the answer of this question below:A man is moving with...
Given :
u = 36 km/h = 36x (5/18) = 10 m/s
Time of reaction, t = 0.9 s
decelration, a = -5 m/s^2
v = 0 m/s
Let s1 be the distance covered by man when the object is seen by him
we have, u = s1/t
because s1 = u x t
= 10 x 0.9
= 9 m
when he applies brakes and decelerates at the rate of 5 m/s^2, the distance covered by him is s2
We have, v^2 = u^2 + 2as2
therefore 0 = u^2 - 2as2 ( therefore v=0 m/s)
therefore s2 = u^2/2a = 10^2/ (2 x 5)
therefore s2 =10 m
So, total distance covered s = s1 + s2 = 9 +10 =19 m
Hence 19 m distance covered by man before he stops.
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Can you explain the answer of this question below:A man is moving with...
Given:
Initial velocity (u) = 36 kmph = 10 m/s
Time of reaction (t1) = 0.9 seconds
Deceleration (a) = 5 m/s2
To find: Total distance covered before stopping
Approach:
We need to find the distance covered during the time of reaction and the distance covered during the deceleration.
1. Distance covered during the time of reaction:
During the time of reaction, the man keeps on moving with the initial velocity. Therefore, the distance covered during this time is given by:
d1 = u * t1
d1 = 10 m/s * 0.9 s = 9 m
2. Distance covered during the deceleration:
After applying the brakes, the man decelerates at 5 m/s2. We know that the distance covered during the deceleration is given by:
d2 = (v^2 - u^2) / 2a
where v is the final velocity, which is 0 m/s (as the man stops) and u is the initial velocity.
d2 = (0^2 - 10^2) / 2 * (-5)
d2 = 100 / 10 = 10 m
3. Total distance covered:
The total distance covered is the sum of distance covered during the time of reaction and distance covered during the deceleration.
Total distance (d) = d1 + d2
d = 9 m + 10 m = 19 m
Therefore, the total distance covered before the man stops is 19 m.
Initial velocity (u) = 36 kmph = 10 m/s
Time of reaction (t1) = 0.9 seconds
Deceleration (a) = 5 m/s2
To find: Total distance covered before stopping
Approach:
We need to find the distance covered during the time of reaction and the distance covered during the deceleration.
1. Distance covered during the time of reaction:
During the time of reaction, the man keeps on moving with the initial velocity. Therefore, the distance covered during this time is given by:
d1 = u * t1
d1 = 10 m/s * 0.9 s = 9 m
2. Distance covered during the deceleration:
After applying the brakes, the man decelerates at 5 m/s2. We know that the distance covered during the deceleration is given by:
d2 = (v^2 - u^2) / 2a
where v is the final velocity, which is 0 m/s (as the man stops) and u is the initial velocity.
d2 = (0^2 - 10^2) / 2 * (-5)
d2 = 100 / 10 = 10 m
3. Total distance covered:
The total distance covered is the sum of distance covered during the time of reaction and distance covered during the deceleration.
Total distance (d) = d1 + d2
d = 9 m + 10 m = 19 m
Therefore, the total distance covered before the man stops is 19 m.
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Can you explain the answer of this question below:A man is moving with 36 kmph. The time of reaction is 0.9 seconds. On seeing an obstacle in the path, he applies brakes and decelerates at 5 m/s2, the total distance covered before he stops is:A:19 mB:17 mC:16 mD:18 mThe answer is a. for Class 9 2025 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about Can you explain the answer of this question below:A man is moving with 36 kmph. The time of reaction is 0.9 seconds. On seeing an obstacle in the path, he applies brakes and decelerates at 5 m/s2, the total distance covered before he stops is:A:19 mB:17 mC:16 mD:18 mThe answer is a. covers all topics & solutions for Class 9 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Can you explain the answer of this question below:A man is moving with 36 kmph. The time of reaction is 0.9 seconds. On seeing an obstacle in the path, he applies brakes and decelerates at 5 m/s2, the total distance covered before he stops is:A:19 mB:17 mC:16 mD:18 mThe answer is a..
Can you explain the answer of this question below:A man is moving with 36 kmph. The time of reaction is 0.9 seconds. On seeing an obstacle in the path, he applies brakes and decelerates at 5 m/s2, the total distance covered before he stops is:A:19 mB:17 mC:16 mD:18 mThe answer is a. for Class 9 2025 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about Can you explain the answer of this question below:A man is moving with 36 kmph. The time of reaction is 0.9 seconds. On seeing an obstacle in the path, he applies brakes and decelerates at 5 m/s2, the total distance covered before he stops is:A:19 mB:17 mC:16 mD:18 mThe answer is a. covers all topics & solutions for Class 9 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Can you explain the answer of this question below:A man is moving with 36 kmph. The time of reaction is 0.9 seconds. On seeing an obstacle in the path, he applies brakes and decelerates at 5 m/s2, the total distance covered before he stops is:A:19 mB:17 mC:16 mD:18 mThe answer is a..
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