A race car driver must average 203.0 km/h over the course of a time trial lasting ten laps. If the first nine?

Question

A race car driver must average 203.0 km/h over the course of a time trial lasting ten laps. If the first nine laps were done at 200.0 km/h, what average speed must be maintained for the last lap? (Answer to the nearest km/h.)
km/h

Answer 1

Let the total distance, in km, of 1 lap be D. The required
speed is calculated
203 km/hr = 10*D/T
T = 10*D/203 km/hr = (10*D/203) hr

The first 9 laps covered a distance of 9*D in time t1.
200 km/hr = 9*D / t1
t1 = 9*D/200 km/hr = (9*D/200) hr

T = t1+t2
T = (9*D/200) hr + t2
Also we found above
T = (10*D/203) hr
So
(10*D/203) hr = (9*D/200) hr + t2
t2 = 10*D/203 - 9*D/200 hr = 0.00426*D hr

So the final lap has to be driven at
Vfinal = D/t2 = D/(.00426*D) = 235 km/hr
________________________________

If leaving the D in there is hard to grasp, let D = 1 km (you could also make it 100 km if you like).

Then total time must be T where
203 km/hr = 10 km/T or
T = (10/203) hr = 0.04826 hr

The time t1 consumed loafing through the first 9 laps
200 km/hr = 9 km / t1
t1 = (9/200) hr = 0.045 hr

So the time remaining if the race is to be finished in a total time of 0.04826 hr
t2 = T - t1 = 0.04826 hr - 0.045 hr = .00426 hr

So the final lap's speed needs to be
1 km / .00426 hr = 235 km/hr

Answer 2

permit d = the area around one lap permit t = the completed time the motive force took for the 1st 9 laps. Then, considering that distance/time = velocity, the value for the 1st 9 laps is 9d/t = 195. Now permit T be the completed time mandatory for all 10 laps. because of the fact the ten-lap conventional could be 198 km/h, 10d/T = 198. The time for the final lap is T - t and so the value for the final lap is d/(T - t). From above, 9d/t = 195, or t = 9d/195. additionally, considering that 10d/T = 198, then T = 10d/198. subsequently T - t = 10d/198 - 9d/195 = d(10/198 - 9/195) (T - t)/d = ((10)(195) - (9)(198)) / ((195)(198)) = (1950 - 1782)/38610 = 168/38610. consequently the value needed for the final lap is d/(T - t) = 38610/168 = 230 km/h, to the closest km/h. Please word- one in each and every of the different solutions skill that the cost of the completed ten laps is in basic terms the common of the ten speeds of each and every of the guy laps and provide an equation wherein the ten-lap velocity (198) is the weighed conventional of 9/10 of 195 and a million/10 of the final lap's velocity. it is not splendid. the cost of the final lap of course could be greater than 195 with the intention to advance the desirable conventional to 198. yet this suggests that the time for the final lap could be under the common of the previous lap circumstances. So the time for the final lap is under a million/10 of the time for the completed ten laps. A weighted conventional consistent with sort of laps, (this is, of distance) would be too small to do the interest. the splendid answer may be the weighted conventional consistent with time. right here's a user-friendly occasion as an occasion this ingredient: think you have a song that's 36 km according to lap, and you prefer to return and forth 2 laps with an conventional velocity of 9 km/h. Now think which you finished the 1st lap on the cost of 6 km/h. you will think of that the 2d lap could have velocity of 12 km/h considering that (6 + 12)/2 = 18/2 = 9. yet one lap at 6 km/h takes 36/6 = 6 hours. One lap at 12 km/h takes 36/12 = 3 hours. so which you have long previous 2 laps = seventy two km in 6 hr. + 3 hr. = 9 hours for an conventional velocity of in basic terms seventy two/9 = 8 km/h. with the intention to conventional 9 km/h you are able to return and forth the 2d lap at 18 km/h. The time for the 2d lap is then 36/18 = 2 hours and th time for the two laps is 6 hr. + 2 hr. = 8 hours, and seventy two km in 8 hours leads to an conventional velocity of seventy two/8 = 9 km/h.

Answer 3

If the last lap's average speed = x
then (200 + x) / 2 = 203
200 + x = 203*2
200 + x = 406
x = 406 - 200
=206
The average speed of the last lap is 206 k.m/h
o.k

Answer 4

Its around 210 km/h

Answer 5

230.0 km/h