# Explain?

• Feb 23rd 2010, 06:44 AM
brianfisher1208
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!
• Feb 23rd 2010, 09:09 AM
icefirez
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
• Feb 23rd 2010, 09:22 AM
Quote:

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.

$distance\ fallen=kt^2$

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

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

After 3 seconds

$d=16.1(3^2)\ feet$
• Feb 23rd 2010, 09:26 AM
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
• Feb 23rd 2010, 09:31 AM
icefirez
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
• Feb 23rd 2010, 09:45 AM
Quote:

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... $r=\frac{k}{t}$

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

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

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

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

$(av\ speed)time=distance$

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

$170(1.5)=x(1.45)$

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

If the racecar takes a shorter time, it's average speed must have increased.
• Feb 23rd 2010, 11:53 AM
Quote:

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....

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

$d=(av.speed)t$

Distance is the same for both laps, so

$170(90)=x(87)$

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