Question

A bear spies some honey and takes off from rest, accelerating at a rate of 2.0 m/s^2. If the honey is 16m away, how fast will his snout be going at the moment of ecstasy?

Answer 1

Answer:

8 m/s

Explanation:

From the third law of motion

$v^{2}=u^{2}+2as$ where v is final velocity, u is initial velocity, a is acceleration, s is distance covered.

Since the bee takes off from rest, initial velocity is zero.

Substituting 0 for u, the equation becomes

$v^{2}=0+2as=2as$ and making v the subject of the formula

$v=\sqrt {2as}$

Substituting 2 for a and 16 m for s

$v=\sqrt {2*2*16}$

v=8 m/s