Thread: Finding an instantaneous rate of change

1. Finding an instantaneous rate of change

What function do you plug in 1/15??? That's the one that really confuses me... Please help

2. Re: Finding an instantaneous rate of change

Originally Posted by asilvester635
What function do you plug in 1/15??? That's the one that really confuses me... Please help
4(min)/60(min = 1 hour) = 1/15 by simplifying, did you get? it is the time t, say the x_2 point, if you were to put those data on the x-axis, is it clear now?

dokrbb

3. Re: Finding an instantaneous rate of change

You don't "plug" the time in to any formula. The problem tells you that "s(t)" is the distance the car has gone, from the first police car, in time, t, in hours. You are told that the car passes the second police car in 1/15 of an hour so that s(0)= 0 (0 hours after passing the first car- i.e when he is at the first police car, his distance from that car is 0 miles) and s(1/15)= 5 (1/15 hour after passing the first police car, he is at the second police car so has gone 5 miles).