Question

It has been claimed that an insect called the froghopper (Philaenus spumarius) is the best jumper in the animal kingdom. This insect can accelerate at 4 000 m/s2 over a distance of 2.0 mm as it straightens its specially designed " jumping legs." Assuming a uniform acceleration, what is the velocity of the insect after it has accelerated through this short distance.

Answer 1

v=4m/s

Step-by-step explanation:

The formulas for accelerated motion are:

`v=v_0+at\\x=x_0+v_0t+(at^2)/(2)`

We can derive the formula
`v^2=v_0^2+2ad` from them.

We have:

`v-v_0=at\\t=(v-v_0)/(a)`

And substitute:

`x=x_0+v_0((v-v_0)/(a))+(a)/(2)((v-v_0)/(a))^2\\x-x_0=(v_0(v-v_0))/(a)+((v-v_0)^2)/(2a)\\2a(x-x_0)=2v_0(v-v_0)+(v-v_0)^2=2v_0v-2v_0^2+v^2+v_0^2-2vv_0=v^2-v_0^2`

Where in the first step of the last row we just multiplied everything by 2a. Since
`x-x_0` is the displacement d, we have proved that
`v^2=v_0^2+2ad`

We use then our values to calculate the final velocity when starting from rest, traveling a distance 0.002m with acceleration
`4000 m/s^2`:

`v=√(v_0^2+2ad)=√(2(4000m/s^2)(0.002m))=4m/s`