# Speed and Time problem

• Aug 7th 2011, 04:08 AM
fleurdel
Speed and Time problem
At 9 a.m., John was cycling from Point A to Point B. At the same time, Mary was cycling from Point B to Point A, using the same route as John. Cycling at 4km/hr faster than Mary, John would pass Mary 300m away from the midpoint.

(a) What time did they pass each other?
(b) If John took another 3 mins to reach Point B, what time would Mary reach Point A?

Thank you for your help.
• Aug 7th 2011, 06:05 AM
Bacterius
Re: Speed and Time problem
Try and visualize it. Since John moves faster than Mary, they will pass each other 300m beyond the midpoint (not before). Thus we get:

J --------------------------- MIDPOINT ----- X ---------------------- M

So this means, John has covered exactly 600m more distance than Mary, just by going 4km/h faster (do you see why? this is the key to solving this problem). Putting this into equations (600m = 0.6km, we are working in km/h here):

$V_{John} = V_{Mary} + 4$
$D_{John} = D_{Mary} + 0.6$

Therefore using $v = \frac{d}{t}$ we get:

$D_{John} = D_{Mary} + 4t$
$D_{John} = D_{Mary} + 0.6$

Which simplifies to $4t = 0.6$, thus $t = 0.15$ (hours). This is 9 minutes, therefore Mary and John will pass each other at 9:09am.

By drawing a diagram, can you see how to solve question (b) now?

EDIT: had made a mistake converting 0.15 to minutes... fixed now.
• Aug 7th 2011, 08:49 AM
fleurdel
Re: Speed and Time problem
Thanks bacterius, I got (a) already can someone show me part (b)?

I got 9:36 am, but the answer is 9:35am
• Aug 8th 2011, 11:13 AM
fleurdel
Re: Speed and Time problem
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