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

8 m/s

Step-by-step 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