# Homework Help: Kinematics motion

1. Dec 6, 2013

### negation

After 35mins of runnin, at the 9km point in a 10km race, you find yourself 100m behind the leader and moving at the same speed.
What should your acceleration be if you're to catch up by the finish line? Assume that the leader maintains constant speed.

Last edited: Dec 6, 2013
2. Dec 6, 2013

3. Dec 6, 2013

### negation

vf = 9000m/35min = 257m min^-1
vf = 9100m/35min = 260 min^-1

10,000-9100m = 900m
900m/ 260m min^-1 = 3.46min

Runner requires 3.46min to tie the leader at the finish line.

xf - xi = 0.5(vf + vi)t
1000m = 0.5(vf+257m min^-1)3.46min
vf = 321m min^-1

vf doesn't tally with the answer sheet

4. Dec 6, 2013

### negation

Can someone help me? I've spent 4 hrs on this question already. I don't want to get into the unhealthy habit of not sleeping for days until I figure out the answer.

5. Dec 6, 2013

### Staff: Mentor

The numbers that are significant in this problem are the 1000 m that the slower runner needs to cover, and the 900 m that the faster runner has left.

Don't use the same variable for both runners:
vf can't possibly be equal to two different numbers.

6. Dec 6, 2013

### CWatters

I agree. You have to get to the finish in less than 3.46min.

Than I would look at one of the SUVAT equations to work out the minimum constant acceleration required. http://en.wikipedia.org/wiki/Equations_of_motion

Perhaps this one and solve for a...

s = ut + 0.5at2

You know..

s = 900m
u = 257m/min or 4.28m/S
t = 3.46min or 207.6 seconds