1. ## Explain?

The speed (r) of a race car around a track varies inversely as the time it takes to go one lap around the track. When Car 44 is traveling 170 mph, it takes him 1.5 mins to complete one lap around the track. If he can complete a lap in 1.45 mins, how fast would he be traveling?

The distance a stone falls when dropped off a cliff is proportional to the square of the time it falls. If the stone falls 64.4 ft in 2 secs, how far would it have fallen in 3 secs?

No entiendo. I don't understand. Thanks for any help!

2. 2)

if stone fall 64.4 ft in 2 sec x ft would be in three seconds

64.4ft : 2 = x : 3
from this we got
2x=3*64.4
now
x=193.2/2
x=96.6ft

3. Originally Posted by brianfisher1208
The speed (r) of a race car around a track varies inversely as the time it takes to go one lap around the track. When Car 44 is traveling 170 mph, it takes him 1.5 mins to complete one lap around the track. If he can complete a lap in 1.45 mins, how fast would he be traveling?

The distance a stone falls when dropped off a cliff is proportional to the square of the time it falls. If the stone falls 64.4 ft in 2 secs, how far would it have fallen in 3 secs?

No entiendo. I don't understand. Thanks for any help!
For 2

The distance is not "linearly" proportional to time,
it's proportional to the square of the time elapsed since release,
as gravity accelerates the stone to the cliff base or water.

$\displaystyle distance\ fallen=kt^2$

$\displaystyle 64.4=k(2^2)=4k$

$\displaystyle k=\frac{64.4}{4}=16.1$

After 3 seconds

$\displaystyle d=16.1(3^2)\ feet$

4. 1) if car traveling 170 mph complete lap for 1.5 or(90 sec) for x speed (mph) he would travel the lap 1.45(87 sec)

170mph : 90 s = x mph : 87 s

90x = 170 * 87

90x = 14790mph*s

x = 14790 mph*s/90s

x = 164.33 mph...

here it is ....this should be fine

5. Archie Meade that was nice...didnt readed well...anyway i cant speak so well english and dont know good math terms ... but i checked now ..by the way thanks...

i dont know if this second answer is correct

6. Originally Posted by brianfisher1208
The speed (r) of a race car around a track varies inversely as the time it takes to go one lap around the track. When Car 44 is traveling 170 mph, it takes him 1.5 mins to complete one lap around the track. If he can complete a lap in 1.45 mins, how fast would he be traveling?
Inverse relationship... $\displaystyle r=\frac{k}{t}$

$\displaystyle 170=\frac{k}{\left(\frac{1.5}{60}\right)}$ by converting minutes to hours.

$\displaystyle k=\frac{170(1.5)}{60}=4.25$

$\displaystyle r=\frac{4.25}{\left(\frac{1.45}{60}\right)}=\frac{ 60(4.25)}{1.45}$ mph

$\displaystyle \frac{distance}{time}=average\ speed$

$\displaystyle (av\ speed)time=distance$

distance = lap, so it's the same both times

$\displaystyle 170(1.5)=x(1.45)$

$\displaystyle x=170\frac{1.5}{1.45}$

If the racecar takes a shorter time, it's average speed must have increased.

7. Originally Posted by icefirez
1) if car traveling 170 mph complete lap for 1.5 or(90 sec) for x speed (mph) he would travel the lap 1.45(87 sec)

170mph : 90 s = x mph : 87 s

90x = 170 * 87

90x = 14790mph*s

x = 14790 mph*s/90s

x = 164.33 mph...

here it is ....this should be fine
Hi icefirez,

you are almost there!
but you left out an important piece at the start....

$\displaystyle av.speed=\frac{d}{t}$

$\displaystyle d=(av.speed)t$

Distance is the same for both laps, so

$\displaystyle 170(90)=x(87)$

$\displaystyle x=\frac{170(90)}{87}$