When jumping, a flea reaches a takeoff speed of 1.0 m/s over a distance of 0.47 mm .What is the flea's acceleration during the j

Question

Gwar

27 days ago

15

When jumping, a flea reaches a takeoff speed of 1.0 m/s over a distance of 0.47 mm .What is the flea's acceleration during the j

ump phase?

Physics

2 answers:

inna [2.2K] · Answer 1 · 2026-02-26T15:24:36+03:00

The acceleration of a flea during its jumping phase is approximately 1100 m/s²

Additional Details

Acceleration refers to the change in velocity over time.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

a = acceleration (m/s²)

v = final velocity (m/s)

u = initial velocity (m/s)

t = time duration (s)

d = distance (m)

Now, let's address the problem!

This question pertains to Kinematics.

We will proceed with the following approach

Provided:

initial velocity = u = 0 m/s

final velocity = v = 1.0 m/s

distance covered = d = 0.47 mm = 4.7 × 10⁻⁴ m

Required:

acceleration = a =?

Solution:

$v^2 = u^2 + 2ad$

$1^2 = 0^2 + 2a(4.7 \times 10^{-4})$

$a = 1 \div (9.4 \times 10^{-4})$

$a = 1063.830 ~ m/s^2$

$a \approx 1100 ~ m/s^2$

Further Learning

Runner's Velocity:
Kinetic Energy:
Understanding Acceleration:
Car Speed:

Answer Summary

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity, Driver, Car, Deceleration, Acceleration, Obstacle, Speed, Time, Rate

Ostrovityanka [2.2K] · Answer 2 · 2026-02-26T15:26:50+03:00

Assuming constant acceleration for simplicity, even if not entirely realistic, we have the final speed, vf, at 1.0 m/s, and we cover a distance, d, of 0.47 mm, which converts to 0.00047 m (as 1 m equals 1000 mm). At the start of its jump, we can set the initial velocity, vi, to be 0. Using the formula vf^2 = vi^2 + 2ad, we solve for acceleration, a, as follows: a = (vf^2 - vi^2)/(2d) = (1.0^2 - 0^2)/(2*0.00047) = 1,064 m/s^2, which is quite impressive for a flea!