Help with a word problem!!!?

Question

A bus traveling at an average rate of 30 miles per hour left the city at 11:45 A.M. A car following the bus at 45 miles per hour left the city at noon. At what time did the car catch up with the bus?

(if u can please just show me how to write the prblem out, ican solve it.... thanx so much!!!)

Answer 1

You could write two equations using distance - speed times time. Since the distances are equal, so are speed x time.

t = hours bus travels
t - 0.25 = hours car travels (since it started a quarter of an hour later than the bus)

30t = 45 (t-0.25)

Answer 2

let t be the time in hours since bus left the station
distance D the bus travelled is 30t

the time the car traveled is [ t-1/4] because 15 minutes is 1/4 an hour
distance car traveled D=45[t-1/4]

the distances are the same when D=D or
30t=45[t-1/4] clear bracket
30t=45t-45/4 subtract 30t from each side of = sign
0=15t -45/4 add 45/4 to each side
45/4 = 15t divide each side by 15
3/4 =t or they meet 45 minutes after the bus left or at 12:30


another way
the car travels 15 mph faster than the bus
the bus traveled 15 minutes[or 1/4 hour] longer than the car or 30/4 miles more
it takes the car [30/4] / 15 hours to catch the bus or 1/2 hour after the car left at 12 o clock
the car reaches the bus at 12:30

Guido

Answer 3

Okay...your basic formula is Distance = (Speed)(Time)

And when both vehicles are the same distance from the city you want to know the time that happens...So the distances will be equal...That is where we start...
Distances are equal...

So

30 (T) = 45 (T - 15) where T is the time traveled

30T = 45T - 675

-15T = - 675
or if you multiply both sides by -1
15T = 675

and T = 675 / 15 = 45 minutes

So 11:45 plus 45 minutes is 12:30 A.M.

Answer 4

the distance between thebus and the cr at noon
=(1/4)*30=30/4 miles
relativespeed =45-30=15 mph
time taken to close the gao of 30/4 miles
=(30/4)*1/15=1/2
so at 12.30 p.m.the car would catch up with the bus

Answer 5

At noon the bus is 30/4 = 7.5 miles from the starting point.

So
time * 30mph + 7.5miles = time*45mph
get time one one side
7.5 = 45 *time - 30*time
subtract the time
7.5 = 15*time
divide by 15
7.5/15 = (15*time)/15

time = .5 hours
which would be 30 minutes from Noon.
The answer should be 12:30

Answer 6

Sure, (I didn't solve it since you didn't want me to)
When they meet they will have travelled the same distance but not in the same time. Velocity = distance/time so d= Vt
Let t be the time (in hours) the bus travels until they meet, then t-0.25 hrs is the time the car travels until they meet, so since the distances must be equal, we must have:

30mph x t hrs = 45mph x (t-0.25)hrs

Now solve for t.

Answer 7

30x=45(x-.75)
30x=45x-33.75
15x=33.75
x=2.25=2 hrs 15 min.
start 11:45AM
caught at 2PM