WEBVTT 00:00:00.000 --> 00:00:03.250 This is College Physics Answers with Shaun Dychko. 00:00:03.500 --> 00:00:06.980 The question here is by what factor does the drag force increase 00:00:06.980 --> 00:00:10.860 when the car increases its speed from 65 kilometers an hour 00:00:10.880 --> 00:00:13.180 to 110 kilometers an hour? 00:00:13.540 --> 00:00:19.120 So the drag force formula is one-half times drag coefficient times density of air times 00:00:19.220 --> 00:00:23.060 the cross-sectional area of the object times its speed squared 00:00:23.160 --> 00:00:26.420 and I put a subscript only on the speed because that's the only 00:00:26.420 --> 00:00:29.060 factor that changes between the two scenarios— 00:00:29.060 --> 00:00:30.660 the first and the second scenario— 00:00:30.660 --> 00:00:33.760 so we have speed 1 and we have speed 2. 00:00:34.040 --> 00:00:36.800 So when we divide these two drag forces, 00:00:36.900 --> 00:00:39.620 all these factors are going to cancel 00:00:39.620 --> 00:00:44.000 and we are left with only v 2 squared divided by v 1 squared 00:00:44.480 --> 00:00:47.500 and that can be written as v 2 over v 1 all squared. 00:00:47.960 --> 00:00:51.960 I noticed that it's not necessary to change the units into meters per second 00:00:51.960 --> 00:00:54.660 although that's normally what we would do to have 00:00:54.660 --> 00:00:57.060 mks units—meters, kilograms and seconds— 00:00:57.060 --> 00:01:00.540 but in this case, the units just have to be the same so that they cancel 00:01:00.540 --> 00:01:03.120 and we'll be left with an answer of 2.9. 00:01:03.200 --> 00:01:05.780 So we can say that the drag force in case two 00:01:05.940 --> 00:01:13.740 is increased by a factor of 2.9 compared to the drag force in scenario one.