Question

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 tailwind 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?

Answer 1

Answer:

Hence Distance traveled at speed 115 mph is 345 miles and Distance traveled at speed 105 mph is 210 miles.

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 345 miles and Distance traveled at speed 105 mph is 210 miles.