Lara rode her motorcycle 300 miles. If she averaged 10mph faster, the trip would have taken her 1 hour less time.

Question

Lara rode her motorcycle 300 miles. If she averaged 10mph faster, the trip would have taken her 1 hour less time. Find Lara's actual speed. Write an equation and solve.with steps 

Answer 1

Trenton has answered the question above. Since I was already working on it, and didn't have a chance to post it until now, I'm submitting my answer as well, since I solved it for speed (vs. for time that Trenton does above, and then substitute for speed), so this is an alternate solution...

 

Let's say Lara rode her motorcycle at X mph, covering the 300 miles in Y hours . Representing this as an equation

X mph * Y hours = distance traveled = 300 miles

If she averaged 10mph faster , her speed is now X + 10 mph

The problem states the trip would have taken her 1 hour less time. So the time she travels now is Y - 1 hour. The distance she travels is still the same, i.e. 300 miles

In other words

(X + 10) mph * (Y-1) hours = 300 miles

So we have 2 equations

X*Y = 300 => Y = 300/X ... result 1

(X+10)*(Y-1) = 300

Substituting Y (time) from result 1 in the above equation,

 

(X+10)*(Y - 1) = 300

gives
(X + 10)*(300/X - 1) = 300

300 + 3000/X - X - 10 = 300

=> X - 3000/X + 10 = 0
Let's multiple both sides by X

X² - 3000 + 10X = 0

(X + 60)(X - 50) = 10

 

Solving it gives X=-60mph or X=50mph

 

Since the speed cannot be negative, the only answer is that Lara's actual speed to start with was 50mph, and she takes Y=300/X = 300/50 = 6 hours to do it initially. 

 

 

Answer 2

Distance = Rate x Time

 

So, our first equation is 300 miles = (speed in mph) x (amount of time traveled). I write this as 300=r*t

 

Now for the second equation, we need to account for the 10mph faster speed (r+10), and for the 1 hour decrease in time traveled (t-1), the distance traveled is still 300. Thus our second equation is:

 

(r+10)*(t-1)=300

 

Solving our second equation for r gives us

 

r=(300/(t-1))-10

 

Plugging this value of r into the first equation, 300=r*t, and solving for t algebraically yields the quadratic equation 10t^2-10t-300=0, which we can factor to 10(t+5)(t-6)=0. This gives us the solutions t=-5 and t=6. However we can disregard t=-5 because -5 hours is not a meaningful amount of time with respect to this problem.

 

So let's take our answer t=6 hours, and plug it into the first equation to find the rate:

 

(300 miles)=(6 hours)*(rate in mph)

 

(300 miles)/(6 hours)=(rate in mph)

 

(300 miles)/(6 hours)=50 mph

 

So, 50 mph is the answer. To check this, we can plug these values for r and t into the second equation:

 

(r+10)*(t-1)=300

(50+10)*(6-1)=300

(60)*(5)=300, which is true so this answer has been double checked!

 

So, in conclusion, Lara went 300 miles in 6 hours at the rate of 50 mph. If she averaged 10 mph faster, Lara would have traveled 300 miles in 5 hours going at the rate of 60 mph.