How fast must he travel for the second half of the trip?
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March 10th, 2011 at 4:28 pm
45mph
March 10th, 2011 at 5:23 pm
as fast as he can..
March 10th, 2011 at 6:17 pm
yeah 45mph right?
March 10th, 2011 at 6:17 pm
45 miles per hour.
Here’s why:
Let x represent his speed for the second half of the trip. The average of his speed for the first half and the second half must be 30, so (15+x)/2 = 30. Then just solve for x: 15 + x = 60, so x = 45.
March 10th, 2011 at 7:07 pm
There’s not enough information given to solve the problem. It would be easy if the truck traveled 15 mph for the first half of the time of a trip, wherein the answer 45 would be solved from:
t(15+x)/(2t)=30 and solving for x yields 45.
However, the wording of this problem suggests that the truck travels half the distance at some time and wants the average speed to be 30 mph so the 2nd half of the distance would be covered at a higher speed at a lesser time. The equation for this can then be expressed as:
(15a+bx)/(a+b) = 30 where a is the time traveled at 15 mph and b is the time traveled at some x speed to average 45 mph.
Solving for x then gives:
x = [(15a)/b] + 30
Similarly, solving for b would give:
b = (15a)/(x-30)
In order to solve this problem one must know either the distance to be traveled (in part or in whole) or some element of time (how long the whole trip took or how long the person drove for 15 mph).
March 10th, 2011 at 7:27 pm
Average rate is equal to total distance divided by total time. If you let d be the first half distance as well as the second half distance, and if you let x be the second half rate, then the time for the first half is d/15 and the time for the second half is d/x. The total time is therefore d/15 + d/x which equals (dx + 15d)/15x. The average rate (which we know to be 30) now equals (2d) divided by (dx + 15d)/15x. When this equation is solved the d’s cancel and the resulting equation is 30(x + 15) = 30x. This equation, of course, has no solution. It appears you need more information, such as a specific distance or a specific time for the first half of the trip.
March 10th, 2011 at 8:00 pm
Infinite. The driver has to double his average and has travelled half the distance, so he used up all the time
which had been needed to do the total distance with the
wanted average speed.
March 10th, 2011 at 8:16 pm
pier_haringsma is correct. You do not need any additional information. Let
D = total distance of the trip
D/2 = the distance the driver has gone so far
T1 = amount of time he took to travel the first half of the trip
T2 = the complete length of time of the whole trip (both halves).
Now, he wants the average ( D / T2 ) to equal 30. His average so far is ( (D/2) / T1 ) = 15.
D/T2 = 30, so D = 30*T2
(D/2) / T1 = 15, so D = 30*T1
D = 30*T1 = 30*T2, so T1 = T2.
So the driver wants the total time for the trip (BOTH halves), T2, to equal the time for the first half of the trip, T1. This means he would have to travel the second half of the trip in no time, which is impossible.