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Andrei [34K]
3 years ago
9

A 555-mile, 5-hour plane trip was flown at two speeds. For the first part of the trip, the average speed was 105 mph. Then the t

ailwind picked up, and the remainder of the trip was flown at an average speed of 115 mph. For how long did the plane fly at each speed?
Mathematics
1 answer:
Pepsi [2]3 years ago
6 0

Answer:

Hence Distance traveled at speed 115 mph is <u>345 miles</u> and Distance traveled at speed 105 mph is <u>210 miles</u>.

Step-by-step explanation:

Given:

Total Distance = 555 mile

Total time = 5 hour

Let time taken to travel at 115 mph be t

So Time taken to travel at 105 mph will be equal to difference between Total time and  time taken to travel at 115 mph.

Time taken to travel at 105 mph = 5-t

Distance traveled in speed 105 mph will be equal to speed multiplied by time taken to travel at 105 mph.

Framing in equation form we get;

Distance traveled in 115 mph = 115t

Also;

Distance traveled in speed 105 mph will be equal to speed multiplied by time taken to travel at 105 mph.

Framing in equation form we get;

Distance traveled in 105 mph = 105(t-5) = 105t - 525

Now Total Distance traveled is equal sum of Distance traveled in 115 mph and Distance traveled in speed 105 mph

115t-(105t-525)=555\\\\115t-105t+525=555\\\\10t+525=555\\\\10t=555-525\\\\10t=30\\\\t=\frac{30}{10}=3

Time taken to travel at 115 mph = 3 hrs.

Time taken to travel at 105 mph = 5-t=5-3=2\ hrs

Now Distance traveled in 115 mph = 115t = 115 \times 3 = 345\ miles

Distance traveled in 105 mph = 105(5-t) = 105(5-3)= 105\times 2 = 210\ miles

Hence Distance traveled at speed 115 mph is <u>345 miles</u> and Distance traveled at speed 105 mph is <u>210 miles</u>.

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