physics ek

[moneyking]

ln many harbors, old automobile tires a re hung along the
sides of wooden docks to cushion them from the impact of
docking boats. The tires deform in accordance with Hooke's
law. As a boat is brought to a stop by gently colliding with
the tires, the rate of deceleration of the boat:
OA) is constant until the boat stops.
OB) decreases until the boat stops.
OC) increases until the boat stops.
O D) increases and then decreases before the boat stops.

Why is the answer C? and not D?
Thanks

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[Cawolf]

Hooke's law is that F = -kx, with k being constant and x being the displacement of the spring.

Since acceleration is directly proportional to force, and force increases with displacement (deformation of the tire), the rate of deceleration increases. This will increase until the boat is fully stopped; at which point the a=0.

If the deceleration were to decrease, as in choice D, you would be violating Hooke's law. The force on the boat from the tire increases with the deformity of the tire, not increases and then decreases. The only way one could argue for choice D is to say that the boat bounces off, and we are looking at the acceleration (not just deceleration). This is not what the question is asking for though.

 

[moneyking]

Thank you for clearing things out!