# word problem involving speed

• Sep 28th 2008, 06:30 PM
yoleven
word problem involving speed
A student is driving home a distance of 150 miles for the weekend. The student knows by experience that increasing the average speed by 10mph, She can reduce the trip time by 35 minutes. What's the actual speed?

D=vt
150=vt

t= 150/v

150=v+10((150/v)-35)
Here, I'm kind of stuck. I was wondering if someone could talk me through this? I'm not sure if I am thinking about this correctly.
• Sep 28th 2008, 07:11 PM
Jhevon
Quote:

Originally Posted by yoleven
A student is driving home a distance of 150 miles for the weekend. The student knows by experience that increasing the average speed by 10mph, She can reduce the trip time by 35 minutes. What's the actual speed?

D=vt
150=vt

t= 150/v

150=v+10((150/v)-35)
Here, I'm kind of stuck. I was wondering if someone could talk me through this? I'm not sure if I am thinking about this correctly.

first note that 35 mins = 7/12 hours. we need our units to be consistent.

Let $t$ be the time it takes to drive 150 miles at a speed of $v$. then we know

$t = \frac {150}v$ .................(1)

if we increase the speed by 10, the time falls by 7/12, so that

$t - \frac 7{12} = \frac {150}{v + 10}$

$\Rightarrow t = \frac {150}{v + 10} + \frac 7{12}$ ................(2)

thus we have two equations, two unknowns. we can solve simultaneously. equate the $t$'s to get

$\frac {150}v = \frac {150}{v + 10} + \frac 7{12}$

multiply through by $12v(v + 10)$, we get:

$12(v + 10) \cdot 150 = 12(150)v + 7v(v + 10)$

can you continue?