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A Car is travelling on a straight road. The maximum velocity the car can attain is 24 m/s . The maximum acceleration and decleration it csn attain are 1 & 4 m/s2 respectively . What will be the shortest time the car takes from rest to rest in a distance of 200 m.

```
2 years ago

Khimraj
3007 Points
```							please check acceleration, decleration, max velocity and distance value and confirm it.…................................................................................................
```
2 years ago
Rajdeep
231 Points
```							HELLO! Maximum velocity that the car can attain is 24 m/s, so v = 24 m/s.Maximum acceleration a = 1 m/s2Maximum retardation = 4 m/s2Distance given = 200m. Car starts from rest and ends by going to rest. Now, time required by the car to start from rest and accelerate to 24 m/s at a rate of 1m/s2  is:v = u + at=> 24 = 0 + 1xt=> t = 24 seconds. Distance covered in this 24 seconds:S = ut + ½ at2=> S = 0 + ½ x 1 x 24 x 24=> S = 288m, which is coming greater than the given value. So, it is proved that there’s something incorrect in the question.The question, hence, is invalid.
```
2 years ago
13 Points
```							Is it 22.36 (10/5).. .. . .. .. ..   .. .. .. ... ... .. .. .. .. .... ... . .. . .. . .. ..  ... .. .. ... .. ... ..  .. ..  ........ ..  ... ... .... ...... ... ....  ..  .. .. .. .. ..  .. .
```
2 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions