00:01
For this problem on the topic of force and motion, we are told that a passenger of mass 85 kilograms is moving along a circular path of radius 3 .5 meters in uniform circular motion.
00:11
We are given the plot of the required magnitude f of the net centipidal force for a range of values of the passenger speed v.
00:20
We want to find the slope of the plot at 8 .3 meters per second.
00:24
We are then shown another plot of the force f for a range of possible values of the period t.
00:30
We want to find the plot slope at 2 .5 seconds.
00:36
So the centripetal force on the passenger is f, which is mv squared over r.
00:45
And the slope of the plot at v is equal to 8 .3 meters per second is the derivative of f with respect to v evaluated at v is 8 .3, which is to mv.
01:03
Over r evaluated at v equals 8 .3.
01:11
And so if we substitute our values in here, this is 2 times 85 kg times the speed 8 .3 meters the second divided by the radius of motion 3 .5 meters...