A-Level Using areas to find distances and displacements

17) The maximum acceleration for a lift is 1m/s/s and the maximum deceleration is 4/s/s. Use a speed-time graph to find the minimum time taken for the lift to go from ground level to the top of a skyscraper 40m high:
[ i ] if the maximum speed is 5m/s
[ ii ] if there is no restriction on the speed of the lift

[ i ] I got that, a trapezium of height 5, takes 5 seconds to reach its max speed, it goes up at 5m/s for t seconds, then takes 1.25 seconds to slow down from its max speed of 5 to 0 at 4m/s/s deceleration.
With that you just solve for t knowing the distance is 40 so minimum time is 11.125s

[ ii ] I can tell its a triangle since it doesnt get capped at 5m/s. t is now the total time. But no clue where to go from there. Answer is 10s in the book

(edited 12 months ago)

Reply 1

old_engineer

Original post by PatrykWalat

17) The maximum acceleration for a lift is 1m/s/s and the maximum deceleration is 4/s/s. Use a speed-time graph to find the minimum time taken for the lift to go from ground level to the top of a skyscraper 40m high:
if the maximum speed is 5m/s
[ii] if there is no restriction on the speed of the lift

I got that, a trapezium of height 5, takes 5 seconds to reach its max speed, it goes up at 5m/s for t seconds, then takes 1.25 seconds to slow down from its max speed of 5 to 0 at 4m/s/s deceleration.
With that you just solve for t knowing the distance is 40 so minimum time is 11.125s

[ii] I can tell its a triangle since it doesnt get capped at 5m/s. t is now the total time. But no clue where to go from there. Answer is 10s in the book

For part (ii) you need to construct a triangle that ramps up with a gradient of 1 then ramps down with a gradient of minus 4, with an area of 40. Try sketching something out and post your working if stuck.