Another motion problem...help, pleeeeaaaasssee?

Question

A car travels 330 miles at a certain speed. If the speed had been 10mph faster, the trip would have been made in one hour less time. Find the speed of the car.

-I'm just starting out on these motion problems and have no one to turn to when confronted with a problem, because I'm doing this course through an 'outlooks' program (i.e. correspondence without conferring with anyone for help). Up until this point, I have had the ability to figure out the material without any assistance, but even using the chart technique explained in the text, which is supposed help clarify the word problems, I'm having difficulties with this section. If anyone could help me both process this problem--seeing as I've got an almost exact duplicate to solve next--and also decide whether or not I should spring for a tutor, I would greatly appreciate the effort.

Answer 1

The equation you need to solve this problem is distance = rate * time.

d = rt

Plug the info you know from the problem into the equation:

d = rt

330= rt     (#1)

This is the equation for when the car travels at the certain speed.

You also need an equation for when the car travels 10 mph faster and takes less time traveling:

So, you know that the distance traveled is still 330 miles.  So, d still equals 330. But now add 10 to r (10 mph faster), and subtract 1 from t (one hour less time):

330 = (r + 10)(t - 1)

Distribute to get:

330 = rt - r + 10t - 10     (#2)

Okay, you have two variables and two equations.  Now, you can solve for those variables.  

Solve for r in equation (#1):

330 = rt

r = 330/t

Plug the value of r into equation (#2) and then solve for t:

330 = 330 - (330/t) + 10t - 10.  Add/subtract like terms:

330 = 320 - (330/t) + 10t.  Subtract 320 from both sides:

10 = -(330/t) + 10t.  Multiply both sides by 10 to cancel out the t in the denominator of (330/t):

10t = -330 + 10t^2.  Subtract 10t from both sides:

0 = 10t^2 - 10t - 330.  Factor out 10:

0 = 10(t^2 - t - 33).  Divide both sides by 10:

0 = t^2 - t - 33.  Use the quadratic equation to solve for t:

t = [-(-1) (+/-) sqrt((-1)^2 - (4 * 1 * -33))] / (2 * 1)

t = [1 (+/-) sqrt(1 + 132)] / 2

t = [1 (+/-) sqrt(133)] / 2

t = [1 + sqrt(133)] / 2,   [1 - sqrt(133)] / 2

t = 6.2663,   -5.2663

Since -5.2663 hours does not make sense, the only answer is 6.2663 hours.

r = 330/t

r = 330 / 6.2663

r = 52.6628 mph

I hope that helped!

And if you're really struggling with math, getting a tutor could greatly help.

Answer 2

Motion problems could be more accurately be called motion sickness.

Some of these things are, in my opinion, meant to improve  the profits

of the liquor industry.

This particular problem you've presented has an untidy answer, not at all

like most motion problems, and so might cause uncertainty in the

solution.

Here's how it is done

..............1st Trip.....2nd Trip

Distance...330...........330

Rate...........R............R+10

Time...........T.............T+1

For 1st Trip, 330=RT [The D=RT formula]

For 2nd Trip, 330=(R+10)(T-1)

If I expand 330=(R+10)(T-1), I get

RT+10T-R-10=330..................Eq.1

From 330=RT, I getR=330/T....Eq.2

Substitute 330/T for R in Eq.1, to get

330=(330/T)T+10T-330/T-10

330=330+10T-330/T-10

10T^2-10T-330=0

T^2-T-33=0

Not easy to factor, and it's here that a student may begin to question his work, because the answer doesn't appear to be "tidy" We press on.

Use the Quadratic Formula.

T={-b+/-rt(b^2-4ac)} / 2a, where

a=1, b= -1, c= -33

T={1+/-rt(1+132)} / 2

T={1+/-rt(133)} / 2

T={1+/-11.53} /2

T=12.53/2 or -10.53/2

We disregard the -10.53/2 solution because Time isn't negative for our

problem. Therefore

T=6.265 hours

Not the usual "tidy" answer

From D=RT, R=D/T

R=330/6.265

R=52.67 mph

Let's check:

D=RT

52.67 X 6.265= 329.98 miles, say 330

We're golden!!

I recommend a tabular set-up like the one I've used. I also draw a bit

of a sketch-just a horizontal line really- and mark off the given

conditions. Helps me see the problem.

The trick then, is to use the D=RT formula to get an equation set-up.

Note that in your problem, we had 2 equations ans 2 unknowns.

We had

330=RT, and

330=(R+10)(T-1)

A lot of students leap at doing this:

RT=(R+10)(T-1), thus producing one equation with 2 unknowns. It

doesn't work, and so get angry at motion problems in general. Only

experience guides you around such a pit-fall.

I sincerely hope this has been of use to you. I hold the opinion that you seem both capable and determined. It's early times yet for you to consider a tutor. Hold off .

I've looked at the other questions you've asked, and by and large you

received good answers to them. When you come across an answerer

whose style you like, why not add them as one of your contacts?

That way, every time you ask a question, they are automatically

alerted and would probably answer.

Sort of your own private tutors!

Cheers