helppppppppp

Hi,Sarah did 12 km (exactly) from 10 am to 11 am - so (I know you got it right as you say), velocity = (v simple maths even for a millennial [who normally automatically reaches for a calculator] - sorry!) 12 kph.

She continued for 3 km (at the same velocity) (3 is a quarter of 12, so it took her a quarter of one hour = 15 minutes extra past her home.

The graph shows that she took a 30 min break from 11 to 11.30 am;

So, the time back home = 1 hour (as on way out plus the 15 min mentioned above = 1 hr 15 min
From 11.30 am if you add this 75 min you get 12.45 pm.

However, this is an ambiguous Q because technically Sarah would have passed her home BEFORE doing the extra 3 km, so the answer should be 15 min less = 12.30 pm (but that would be too simple, I suppose!!) I would guess that if they are as incompetent in physics as they are in biology, THIS IS AN EDEXCEL Q [do let me know, pls!)

M.

this is edexel exactly...but how did you know the speed remains 12kph,,,,I calculated the gradient of the line with negative gradient and find the time, and I got 12:42.

I agree with Wuxinyu - the return speed is 10 km/h ( v = (12 - -3)/(1.5 - 3) = - 10 km/h while speed from home is 12km/h. This means girl passes home at 12:42 pm but what does the question ask?
It doesn’t ask you to calculate the time when girl passes home - it asks you to estimate when this oocurs. You can do this by reading the answer from the graph (12:45 pm).

I see, so this means just estimate what the time is, instead of calculating them...thank u