Question

I really need help with this quastion a jet leaves an airport traveling at a steady rate of 600km another jet leaves the same airport 3/4 later traveling at 800km in the same direction. how long will it take the second jet to overtake the first

Answer 1

3/4? i m assuming it is 3/4hrs
let v is the velocity first jet = 600km/hr
and u is the velocity of second jet =800km/hr
when second jet leave the airport the first is already covered a distance AB =600*3/4
AB= 450
relative velocity of second jet with respect first is V = v-u=200km/hr
the time taken by second to overtake first is =AB/V
T=450/200 =9/4hrs

Answer 2

d=r*t

d=800t => first equation

d-450=600t=> second equation

800t-450=600t
200t=450
t=9/4

d=800=9/4
d=1800 km

So, it will take 2 hours and 15 min to take over the first

Answer 3

Good job aroub, with clear steps.

Answer 4

Oh wait this d=800=9/4 is suppose to be d=800(9/4)

Hehe, thank you radar!! =DD

Answer 5

Yes, but the key is both planes will have traveled the same distance. The problem was only asking how long would the faster plane caught up with slower one and 9/4 hr (assuming the speeds were per hour)

Answer 6

As I said "good answer"

Answer 7

extra information =)

Answer 8

No wait

Answer 9

so the distance is wrong?

Answer 10

The distance would of been the fast plane speed times 9/4 (wouldn't it)

Answer 11

aroud urs distance and time both are correct

Answer 12

*aroub