When jumping, a flea rapidly extends its legs, reaching a takeoff speed of 1.0 m/s over a distance of 0.50 mm . part a what is t

Question

densk

3 months ago

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When jumping, a flea rapidly extends its legs, reaching a takeoff speed of 1.0 m/s over a distance of 0.50 mm . part a what is t

he flea's acceleration as it extends its legs?

Mathematics

2 answers:

tester [12.3K] · Answer 1 · 2026-02-19T12:23:02+03:00

The flea's acceleration during leg extension is $\fbox{\begin\\\ 1000\ m/s^{2}\\\end{minispace}}$ .

Explanation:

Given data:

The flea’s takeoff velocity is $1.0\ \text{m/s}$ .

The flea covers a distance of $0.50\ \text{m/s}$ .

Required:

Determine the acceleration of the flea as its legs push off.

Applied formula:

Acceleration can be calculated using:

$\fbox{\begin\\\ \math v^{2}=u^{2}+2as\\\end{minisapce}}$ ...(1)

Where $v$ represents the flea’s final velocity, $u$ its initial velocity, $a$ the acceleration, and $s$ the distance traveled.

Calculations:

The flea starts at an initial velocity of $0$ .

The final velocity reached is $1.0\text{m/s}$ .

The displacement is originally given in millimeters.

Convert millimeters to meters:

$\fbox{\begin\\\ \begin{aligned}0.50 \text{ mm}&=\dfrac{0.50}{1000}\text{ m}\\&=0.0005\text{ m}\end{aligned}\\\end{minispace}}$

Plugging in values where $0$ replaces $u$ , $1.0$ replaces $v$ , and $0.0005$ substitutes $s$ into equation (1) gives the acceleration.

$\begin{aligned}1^{2}&=0^{2}+(2a\times(0.0005))\\1&=0.001a\\a&=\dfrac{1}{0.001}\\a&=1000\end{aligned}$

Therefore, the flea's acceleration as it extends its legs is $\fbox{\begin\\\ 1000\ m/s^{2}\\\end{minispace}}$ .

Learn more:

1. Linear equation solutions: brainly.com/question/1682776

2. Acceleration magnitude: brainly.com/question/1597065

Answer info:

Grade: Middle school

Subject: Mathematics

Chapter: Speed and distance

Keywords: flea, leap, leg extension, takeoff speed, acceleration, distance covered, 0.05 mm, 1000 m/s², initial velocity, final velocity

zzz [12.3K] · Answer 2 · 2026-02-19T12:26:28+03:00

The flea experiences an acceleration of 1,000 m/s².

Detailed explanation

This scenario involves motion with constant acceleration.

The relevant variables include the following.

$\boxed{u \ or \ v_i = initial \ velocity}$

$\boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}$

$\boxed{a = acceleration \ (constant)}$

$\boxed{d = distance \ travelled}$

We know the flea attains a takeoff velocity of 1.0 m/s over a distance of 0.50 mm.

The flea starts from rest, so the initial velocity is zero. The question asks for the flea's acceleration during leg extension.

The equation we apply is:

$\boxed{ \ v^2 = u^2 + 2ad \ }$

a = acceleration (m/s²)
u = initial speed = 0 m/s
v = takeoff velocity = 1.0 m/s
d = displacement = 0.50 mm

First, convert 0.50 mm to meters: $\boxed{0.50 \ mm = 0.50 \times 10^{-3} \ m = 5.0 \times 10^{-4} \ m}$

Solution steps:

Rearrange the formula to isolate acceleration (a).

$\boxed{ \ v^2 = u^2 + 2ad \ }$

$\boxed{ \ v^2 - u^2 = 2ad \ }$

$\boxed{ \ 2ad = v^2 - u^2 \ }$

$\boxed{ \ a = \frac{v^2 - u^2}{2d} \ }$

Insert the given values into the rearranged equation.

$\boxed{ \ a = \frac{(1.0)^2 - (0)^2}{2(5.0 \times 10^{-4})} \ }$

$\boxed{ \ a = \frac{1}{10 \times 10^{-4}} \ }$

$\boxed{ \ a = \frac{1}{1 \times 10^{-3}} \ }$

$\boxed{ \ a = 1 \times 10^{3}\ = 1,000 \ m/s^2 \ }$

This yields the flea's acceleration as 1,000 m/s².

Additional resources

Scientific notation: brainly.com/question/7263463
Determining substance mass: brainly.com/question/4053884
Conversion into cubic units: brainly.com/question/1446243

Keywords: flea, jumping, leg extension, takeoff velocity, initial speed, displacement, acceleration, constant acceleration, unit conversion