# Homework Help: Motion problem for constant acceleration of rabbit

1. Sep 19, 2010

### chroncile

1. The problem statement, all variables and given/known data
A rabbit runs past a bush where a wolf is lying in wait. The rabbit is running at 3.0 m/s. The wolf leaps out 0.50 seconds after the rabbit has passed and accelerates at 1.2 m/s2 chasing after the rabbit. If the wolf can keep up this acceleration, how long will it take for the wolf to catch the rabbit?

2. Relevant equations
d = vi * t + 0.5 * a * t^2

3. The attempt at a solution
I got 5.85 s, but the answer is 5.46 s so I did it wrong.

2. Sep 19, 2010

### Staff: Mentor

When did you measure the time from? Hint: Where is the rabbit when the wolf starts chasing?

3. Sep 19, 2010

### chroncile

tR = tw + 0.5 s
dw = dr + 1.5 m

Wolf:

vi = 0
a = 1.2 m/s2

d = vit + 0.5 at2
d = 0.5(1.2)t2
d = 0.6t2

Rabbit:

v = 3.0 m/s
d = v * t
d = 3.0t

Since tr = tw + 0.5 s
dr = 3.0(tw + 0.5)
dr = 3.0tw + 1.5

Wolf:

Since dw = dr + 1.5

0.6t2 = 3.0t + 1.5 + 1.5
0.6t2 - 3.0t - 3.0

t = 5.85 or -0.85

4. Sep 19, 2010

### Staff: Mentor

This is true.
This is only true at the moment the wolf starts moving. Skip it.

Good. This is the wolf's position measured from his starting point.

Good. This is the rabbit's position measured from the wolf's starting point.

Nah. Just set the two positions equal to solve for when the wolf reaches the rabbit.

5. Sep 19, 2010

### chroncile

Okay, great I got the right answer, but can you please explain to me why it's not + 1.5?

6. Sep 19, 2010

### Staff: Mentor

Initially the distance between them is 1.5 m. But the wolf catches up to the rabbit, which means they are at the same place. You've already included that head start when you wrote your equations.

The way I would solve it (equivalent to yours of course), measuring everything from the moment and position of the wolf when he starts chasing:

Xr = 3t + 1.5

Xw = 0.6t^2

Just set them equal.