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.