Question

When jumping, a flea rapidly extends its legs, reaching a takeoff speed of 1.0 m/s over a distance of 0.50 mm find acceleration.How long does it take the flea to leave the ground after it begins pushing off?

Answer 1

The acceleration of the flea during takeoff is 1000 m/s², and it takes 0.001 seconds for the flea to reach the takeoff speed and leave the ground using kinematic equations.

Step-by-step explanation:

To find the acceleration of the flea as it jumps, we will use the kinematic equation which relates initial velocity, final velocity, acceleration, and distance covered:

v2 = u2 + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance. The initial velocity (u) is 0 m/s, the final velocity (v) is 1.0 m/s, and the distance (s) is 0.50 mm or 0.50 x 10-3 m.

We rearrange the equation to solve for acceleration (a):

a = (v2 - u2) / (2s)

Substitute the values into the equation:

a = (1.0 m/s)2 / (2 x 0.50 x 10-3 m)

Calculating this gives us:

a = 1.0 m2/s2 / (1.0 x 10-3 m) = 1000 m/s2

Now, we will calculate the time it takes for the flea to leave the ground using the equation of motion:

v = u + at

Since initial velocity (u) is 0 and we have calculated acceleration (a), we can solve for time (t):

t = v / a
= 1.0 m/s / 1000 m/s2
= 0.001 s

The flea takes 0.001 seconds to reach the takeoff speed and leave the ground.

Answer 2

In this question we have given,

initial velocity,u = 0

final velocity, v = 1m/s

displacement =.5mm (Because 1mm=.001m)

We have to find

1.acceleration

2. Time

1. we know third equation of motion is given as,

`v^(2) =u^(2) + 2as`............(1)

Put values of v,u and s in equation (1)

`[tex]s=1^(2) = 0^(2) +2X.0005Xa\\1= .001a\\therefore, a = 1/.001\\ a= 1000m/s^(2)`

2. We know First equation of motion is given as

`v=u+at`......................(2)

put values of v, u and a in equation(2)

`1= 0 + 1000t\\t= 1/1000\\t= 0.001s`