Motion problem in mathematics?

k = m+10

354 miles × 1 hour/k miles = 354/k hours

294 miles × 1 hour/m miles = 294/m hours

354/k = 294/m

354m = 294k

354m = 294(m+10)

60m = 2940

m = 49 miles/hour

k = 59 miles/hour

Kirk averages 59 miles/hour

let mia's speed be x, kirks speed =x+10.

given for time ' t ' distance travelled by kirk= 354 & by mia 294.

hence kirk speed =dist/time; 354/t. =x+10 1

mia's speed =294/t = x 2

1/2 gives

354/294 = (x+10)/x = 1+ 10/x

or 10/49 = 10/x;

x= 49= mia's speed

kirks = 59

If Mia's speed is x than Kirks is x + 10

distance = speed x time or time=distance/speed

Since both took same time then

Mia's time 294/x = 354/(x+10) kirks time

294(x+10) = 354x

294x + 2940 = 354x

2940 = 60x

2940/60 = x just divide to find x

Let the speed of M is a km/h,thus the speed of K will be (a + 10) km/h

Let the time is t hour,thus :-

(a +10) x t = 354-------------(i)

& a x t = 294----------------------(ii)

Now at + 10t = 354 by eq (i)

=>294 +10t = 354

=>10t = 354 - 294 = 60

=>t = 60/10 = 6 hour

by putting this value in eq (i)

=>(a + 10) x 6 = 354

=>(a + 10) = 354/6 = 59 km/h is the required answer.

in some time (we will figure it) Kirk drove 354-294 = 60 miles farther.

Since Kirk had been going 10 mph faster, they must have been driving for 60miles / 10mph = 6 hours.

Now just divide Kirk's distance by the time elapsed. 354miles / 6hrs = his average speed in mph.

For InSearch (2 solutions under): i presumed the airplane velocity contains the wind velocity, consequently with the wind at the back of it, it replaced into going 240 km/hr, no longer 240 km/hr + Wind km/hr. interpreting the question extra heavily, it does say the "average air velocity", so I agree collectively with your calculation InSearch, 40 8 km/hr.

speed=Distance/time