A bear spies some honey and takes off from rest, accelerating at a rate of 1.9 m/s^2. If the honey is 17m away, How fast will the bear be going when he gets to the honey?

17 = (1/2) 1.9 t^2

solve for t
v = 1.9 t

To find the speed of the bear when it reaches the honey, we can use the equation of motion:

v^2 = u^2 + 2as

Where:
v = final velocity (speed of the bear when it reaches the honey)
u = initial velocity (speed of the bear when it starts from rest, which is 0 m/s)
a = acceleration (1.9 m/s^2)
s = displacement (the distance to the honey, which is 17 m)

Plugging in the given values into the equation, we have:

v^2 = 0^2 + 2 * 1.9 * 17

Simplifying this equation gives:

v^2 = 0 + 64.6

v^2 = 64.6

To find the value of v, we need to take the square root of both sides of the equation:

v = √(64.6)

Using a calculator, we can find that:

v ≈ 8.04 m/s

Therefore, the bear will be going approximately 8.04 m/s when it reaches the honey.