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.

Question

Khimraj · Answer

please check acceleration, decleration, max velocity and distance value and confirm it. …................................................................................................

Rajdeep · Answer

HELLO! Maximum velocity that the car can attain is 24 m/s, so v = 24 m/s . Maximum acceleration a = 1 m/s 2 Maximum retardation = 4 m/s 2 Distance 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/s 2 is: v = u + at => 24 = 0 + 1xt => t = 24 seconds. Distance covered in this 24 seconds: S = ut + ½ at 2 => 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.

Advaith · Answer

Is it 22.36 (10/5) .. .. . .. .. .. .. .. .. ... ... .. .. .. .. .... ... . .. . .. . .. .. ... .. .. ... .. ... .. .. .. ........ .. ... ... .... ...... ... .... .. .. .. .. .. .. .. .

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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.

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.

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