Motion problem for constant acceleration of rabbit

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A rabbit runs at 3.0 m/s, and a wolf accelerates at 1.2 m/s², starting 0.5 seconds after the rabbit. The wolf's position equation is derived as d = 0.6t², while the rabbit's position is d = 3.0t + 1.5. To find when the wolf catches the rabbit, their positions are set equal, leading to the equation 0.6t² = 3.0t + 1.5. The correct solution indicates the wolf catches the rabbit in approximately 5.46 seconds, correcting the initial miscalculation of 5.85 seconds. Understanding the initial distance and timing is crucial for solving motion problems involving constant acceleration.
chroncile
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Homework Statement


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?


Homework Equations


d = vi * t + 0.5 * a * t^2


The Attempt at a Solution


I got 5.85 s, but the answer is 5.46 s so I did it wrong.
 
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When did you measure the time from? Hint: Where is the rabbit when the wolf starts chasing?

Show how you got your answer.
 
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
 
chroncile said:
tR = tw + 0.5 s
This is true.
dw = dr + 1.5 m
This is only true at the moment the wolf starts moving. Skip it.

Wolf:

vi = 0
a = 1.2 m/s2

d = vit + 0.5 at2
d = 0.5(1.2)t2
d = 0.6t2
Good. This is the wolf's position measured from his starting point.

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

Wolf:

Since dw = dr + 1.5
Nah. Just set the two positions equal to solve for when the wolf reaches the rabbit.
 
Okay, great I got the right answer, but can you please explain to me why it's not + 1.5?
 
chroncile said:
Okay, great I got the right answer, but can you please explain to me why it's not + 1.5?
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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