   Chapter 2, Problem 61AP

Chapter
Section
Textbook Problem

An insect called the froghopper (Philaenus spumarius) has been called the best juniper in the animal kingdom. This insect can accelerate at over 4.0 × 103 m/s2 during a displacement of 2.0 mm as it straightens its specially equipped “jumping legs.” (a) Assuming uniform acceleration, what is the insect's speed after it has accelerated through this short distance? (b) How long does it take to reach that speed? (c.) How high could the insect jump if air resistance could be ignored? Note that, the actual height obtained is about 0.70 m, so air resistance is important here.

(a)

To determine
The insect’s velocity after straightening its legs.

Explanation

Given Info: The initial velocity of the insect is 0 , the acceleration of the insect is 4000m/s2 , and the displacement of the insect is 2.0×103m .

Explanation:

The formula used to calculate the insect’s velocity after straightening its legs is,

v=v02+2aΔy

Here,

Δy is the displacement of the insect

v is time final velocity of the insect

v0 is the initial velocity of the insect

a is the acceleration of the insect

Substitute 2

(b)

To determine
The time to reach the final velocity.

(c)

To determine
The upward displacement of the insect between when its feet leave the ground and it comes to rest momentarily at maximum altitude.

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