Question

Due to inclement weather, the pilot of a plane slows down the plane’s regular flying rate by 25%. This result to an additional 1.5 hours in covering the 3,000km distance to its regular time required for the trip. Find the regular rate of the trip.

Answer 1

Answer:

Normal speed = 711.11 km/h (Approx)

Step-by-step explanation:

Assume;

Normal speed = x

New speed = x - 25% of x = 0.75 x

Time taken = t

New time taken = t + 1.5 hour

Total distance

Computation:

Total distance = 3,200 km

Speed = Distance/Time

x = 3200 / t

t = 3200/x

so,

0.75 x = 3200 / (t+1.5)

0.75 x = 3200 / (3200/x+1.5)

0.75 x = 3200 / [(3200 + 1,5x)x]

x = 711.11 km/h

Normal speed = 711.11 km/h (Approx)

Answer 2

The regular speed or rate of the trip is 2111.1 km per hour.

Suppose that ; speed of plane is =  S

Due to inclement weather the pilot slow down the plan rate by 25%

then new speed = S-25%  of S = 0.75S

Time taken = t

New time taken = t+1.5 hours

Total distance covered = 3000 km

$\rm{Speed = Distance/Time}$

Initial speed of the plane was,

$S=\dfrac{3000}{t}$

Time t can be written as,

$t=\dfrac{3000}{S}$

So, the new speed will be,

$\begin{aligned}S' &= \dfrac{3000}{t+1.5} \\0.75S &= \dfrac{3000}{t+1.5} \end{aligned}$

So, the regular speed of the trip can be calculated as,

$0.75S = \dfrac{3000}{t+1.5} \\0.75S = \dfrac{3000}{\dfrac{3000}{S}+1.5} \\0.75S = \dfrac{3000S}{3000+1.5S}\\0.75S(3000+1.5S)=3000S$

Solving further,

$625S+1.125S^2=3000S\\1.125S-2375=0\\S=\dfrac{2375}{1.125}\\S=2111.1$

Therefore, the regular speed or rate of the trip is 2111.1 km per hour.

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https://brainly.com/question/23774048