Asleep At The Wheel

Let the distance covered during each lap be $d$ and the speed during second lap be $v$ .

$Speed = \frac{distance}{time}$ and $Time = \frac{distance}{speed}$

Speed during first lap is $90 mph$ . So, time taken for first lap is $\frac{d}{90}$ .

Similarly, time taken for second lap is $\frac{d}{x}$ .

Now, $Average speed = \frac{ Total distance travelled}{Total time taken}$ .

So, $179 = \frac{d+d}{ \frac{d}{90} + \frac{d}{x}}$ .

$x = \boxed{16110}$

To everyone who says the answer should be 268: If the car was traveling at 268 mph for the second lap, it would spend a shorter time traveling at 268 mph than it spent traveling at 90 mph. So, the 268 "contributes" less to the average because you spend less time traveling at 268 mph, so the average is shifted more towards 90 and is less than the 179 you might expect.

Concept is : Total time = Total distance / average speed Let speed in second lap = v size of track is = x time taken in first lap = x/90 time taken in 2nd lap = x/v total time taken = 2x/179 equate the values ==> (x/90)+(x/v) = 2x/179 v = 16110

Ave. Speed= 2(s1×s2)/s1+s2 , where s1= first speed and s2= second speed S1= 90 mph & ave. Speed= 179mph, Substituting known values yields to S2 = 16110mph.

2(90)(x)/(90+x) = 179 solving we get 16110

just use the formula av.spped=2v1v2/v1+v2

Actually I thought my solution is wrong. Why he could drive with a velocity too small like that while the care can reach 16110 mph.