Question

ASAP pls

A train goes from Springfield, MA to Boston at a speed of 40 mph. On the return journey it moves at a speed of 80 mph. If the return journey takes an hour less than the outward journey, what is the average speed overall of the train?

Answer 1

Step-by-step explanation:

u=40mph/u=64.37 km/h

v=80mph/u=128.74km/h

because (1 mile=1.60925km)

so now put the value in average speed formula-

average speed=2uv km/ h

u+v

=2x64.37x128.74

64.37+128.74

=16576.9876

193.11

=85.84 km/h

Answer 2

Answer:

Required average speed is 53.33 m/h.

Step-by-step explanation:

Given,

A train goes from Springfield, MA to Boston at a speed of 40 m/h.

So,in the first case speed of the train is 40 m/h.

Again on the return journey it moves at a speed of 80 m/h.

So,in the second case speed of the train is 80 m/h.

Now we want to find average speed.

But now question is what is average speed.How it will calculate.

This is a problem of speed and distance.

If any object covers distance a to b at a speed of x and the same object covers distance b to a at a speed of y.Then average speed$= \frac{2xy}{x + y}$

Here by given information,

$x = 40$m/h and $y = 80$m/h.

So required average speed$= \frac{2 \times 40 \times 80}{40 + 80} = \frac{2 \times 40 \times 80}{120} = \frac{160}{3} = 53.33$m/h.

Therefore average speed overall of the train is 53.33 m/h.

Here applied formula:-Average speed =$\frac{2xy}{x + y}$

One more important formula of speed and distance,

$Distance = Speed \times Time$