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

Answer:

v=4m/s

Explanation:

The formulas for accelerated motion are:

$v=v_0+at\\x=x_0+v_0t+\frac{at^2}{2}$

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

We have:

$v-v_0=at\\t=\frac{v-v_0}{a}$

And substitute:

$x=x_0+v_0(\frac{v-v_0}{a})+\frac{a}{2}(\frac{v-v_0}{a})^2\\x-x_0=\frac{v_0(v-v_0)}{a}+\frac{(v-v_0)^2}{2a}\\2a(x-x_0)=2v_0(v-v_0)+(v-v_0)^2=2v_0v-2v_0^2+v^2+v_0^2-2vv_0=v^2-v_0^2$

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 $4000 m/s^2$:

$v=\sqrt{v_0^2+2ad}=\sqrt{2(4000m/s^2)(0.002m)}=4m/s$