Question 621569
After receiving your note I checked the problem you referred to and found I had made a mistake, getting the car and the train mixed up, I will correct that here.

Train leaves at 9 am at a speed of 65km/hr straight.
 At 10:30 am a car leaves the same station at 115km/hr.
:
 A) How far from the station does the car catch up with the train?
Let t = travel time of the train
From the information given we know
(t-1.5) = travel time of the car (left an hour and half later)
When the car catches the train, they will have traveled the same distance
Write a distance equation; dist = speed * time
115(t-1.5) = 65t
115t - 172.5 = 65t
115t - 65t = 172.5
50t = 172.5
t = 172.5/50
t = 3.45 hrs is the travel time of the TRAIN
therefore
3.45 * 65 = 224.25 km
Check using the CAR (travel time 1.5 hrs less)
1.95 * 115 = 224.25
:
B) What was the cars average velocity in meters per second?
Find the no. of meters traveled in one hr, divide by 3600 sec
{{{(115*1000)/3600}}} = 31.94 m/sec
:
C) If the car had not left the train station until 11am and the train still travels at 65km/hr, how fast would the car have to go to catch up to the train at the same distance from the station as in part A in Km/hr?
:
let s = speed of the car to accomplish this
travel time of the car is now; 3.45 - 2 = 1.45 hrs
1.45s = 224.25
s = 224.25/1.45
s = 154.655 km/hr to catch the train in 1.45 hrs