JON'S QUESTION IS:
A car is to be brought to the 4th floor of a parking garage(14 m above the ground) by elevator.
Maximum accl. : 0.2 m/s2
Maximum decl. : 0.1 m/s2
Maximum speed reachable : 2.5 m/s
Find the shortest time to make the lift starting from rest and ending at rest.
ANS : I believe you all know this:
When a body accelerates at a constant rate, 'a' from rest for sometime, then moves with constant velocity, 'v' for sometime and then retards at a constant rate, 'r' and comes to rest covering a total distance of 'x', then the total time, 't' for the whole journey is given by :
t = (x / v) + (v / 2)(1/a + 1/r)
SO : Here t = (14 / 2.5) + (2.5 / 2)(1/0.2 + 1/0.1)
= 5.6 + 18.75
= 24.35 seconds
If there is any mistake, please let me know
If right, rate me kya??
Thathwamasi
Moderation Team
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