Uniform Motion Problems.. please help?

Question

Zane began the journey at 5 AM at 30 kilometers per hour. Tricia began to chase him at 7 AM at 40 kilometers per hour. What time was it when Tricia got within 20 kilometers of Zane?

If someone could help me to understand this that would be absolutely wonderful.

Thank you so much in advance.

Answer 1

For both, let t=0 at 5 AM

For Zane,

x(t) = speed * t = 30t

For Tricia,

x(t) = speed * (t - 2 hours) = 40 t - 40*2 = 40t - 80

Since we're solving for when Tricia is 20 kilometers behind...

x(t) (Zane) = x(t) (Tricia) + 20
30t = 40t - 80 + 20
30t - 40t = 40t - 40t -60
-10t = -60
t = 6 hours

Since t = 0 at 5 AM, add 6 hours to that. Tricia is 20 km behind Zane at 11 AM.

Answer 2

This is simple :) Every hour, Tricia gets 10 kilometers closer to Zane. First, you must know Zane has 60 miles at 7 am, and Tricia has 0. At 8 AM, Zane has 90, and Tricia has 40. At 9 AM, Zane has 120, and Tricia has 80. At 10PM, Zand is at 150, and Tricia at 120. Finally, at 11 AM, Zane is at 180, and Tricia at 160. YOu could've found it out by dividing 40 by 10, with 40 as the miles Tricia needed to catch up at the start by the extra distance covered per hour, but this way's easier to get =)

Answer 3

At 7AM, Zane is 60 kilometers ahead of Tricia. Since Tricia is moving 10 kilometers an hour faster than Tricia, she gains 10 kilometers an hour. So at 8AM Zane is 50 kilometers ahead, 9AM 40 kilometers, etc. until 11AM, where Zane is 20 kilometers ahead.

Answer 4

Suppose that after x hrs of travel
Tricia will be 20 km behind Zane
Dist travelled by Zane=30x
Dist travelled by Tricia=40[x-2]
We have the eqn.
30x-40[x-2]=20
-10x=-60
x=6 hrs
That is at 11 AM

Answer 5

the relative speed ofTricia=10kph
thedistance to be closed=20 less than Zane's headstart
=60-20=40 km
time=distance/speed=40/10=4 hours
so Triciawill be within 20 km after 11 AM

Answer 6

I like to write equations for Zane and Tricia

start time at 5 AM, ie. t=0

then
z(t) = 30t
T(t) = 40(t-2) ,

Then you want to find t when

z(t) - T(t) = 20

Subsitute and solve for t:

30t -40t+80 = 20
60 = 10t
6 = t

Now check
z(6) = 180
T(6) = 160

The difference is 20, like you wanted.

This works in more dimensions too!

You can write z(t) = (zx(t), zy(t), zz(t)) if Zane is moving in 3 dimensions.