Question

The velocity v of an elevator moving upward between adjacent floors is shown as a function of time t in the graph above. At which of the following times is the force exerted by the elevator floor on a passenger the least?

I know the answer is E, 6 seconds. Can you explain why?

Answer 1

hint: What does the slope of the tangent line for any point on the graph give you?

Answer 2

Acceleration. So, the point with the steepest tangent line slope aka greatest acceleration has the least force?

Answer 3

No be careful....Newtons second law says F = ma...so force and acceleration are directly related...
as one goes up so does the other

Answer 4

Okay so therefore the lowest acceleration would have the least amount of force. Yes?

Answer 5

Yes...now remember that if an object is decelerating we say the acceleration is negative.
Look at you graph and you can see where the acceleration is negative, zero and positive

Answer 6

So, what is going on at 6 seconds...is the acceleration negative, positive or zero

Answer 7

yes, the force would be the least when the elevator is decelerating the most

Answer 8

So where is the acceleration the most negative?

Answer 9

At 6 seconds. My initial answer was 5 seconds because I was thinking of the initial downward motion of the elevator but I guess that was reading into it too much haha

Answer 10

good

Answer 11

Thank you :)