A bear running initial speed of 4m/s accelerates uniformly for 18s. What is the bears max speed if it travels 135m during the 18s of acceleration
average speed=135/18
average=(Vf+Vi)/2
set them equal, solve for Vf
11 m/s
To find the bear's maximum speed, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (maximum speed)
u = initial velocity (4 m/s)
a = acceleration
s = distance traveled during acceleration (135 m)
First, let's find the acceleration (a):
We can use the formula a = (v - u) / t, where t is the time taken (18 s).
Substituting the given values:
a = (v - 4) / 18
Next, we can substitute the value of acceleration (a) into the distance formula: s = ut + 1/2at^2
135 = 4*18 + 1/2 * a * 18^2
By multiplying and rearranging, we get:
135 = 72 + 162a
Rearranging further:
162a = 135 - 72
162a = 63
a = 63 / 162
a = 0.3889 m/s^2 (approximately)
Now, substituting the value of acceleration (0.3889 m/s^2) into the formula v^2 = u^2 + 2as:
v^2 = 4^2 + 2 * 0.3889 * 135
v^2 = 16 + 106
v^2 = 122
Taking the square root of both sides to solve for v:
v = √122
v ≈ 11.0457 m/s
Therefore, the bear's maximum speed is approximately 11.0 m/s (rounded to one decimal place).