An object moves with acceleration given by a = -6t^3 m/s^2. Its initial velocity at t = 0 is 5 m/s. How far will it move before stopping?
use v^2=Vo^2 + 2*a*x
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To find the distance the object will move before stopping, we need to integrate the equation for acceleration with respect to time to obtain the equation for velocity. Then, we can integrate the equation for velocity with respect to time to obtain the equation for displacement.
Given:
Acceleration, a = -6t^3 m/s^2,
Initial velocity, v₀ = 5 m/s.
First, let's find the equation for velocity (v):
Using integration, we integrate the acceleration equation with respect to time (t) to obtain the equation for velocity:
∫a dt = ∫(-6t^3) dt
Integrating, we get:
v = - (3/4) t^4 + C
To find the constant of integration, C, we use the information that when t = 0, v = v₀:
5 = - (3/4) (0)^4 + C
C = 5
Therefore, the equation for velocity becomes:
v = - (3/4) t^4 + 5
Next, let's find the equation for displacement (s):
Using integration again, we integrate the velocity equation with respect to time (t) to obtain the equation for displacement:
∫v dt = ∫[ -(3/4) t^4 + 5 ] dt
Integrating, we get:
s = - (3/20) t^5 + 5t + D
To find the constant of integration, D, we use the information that when t = 0, s = 0:
0 = - (3/20) (0)^5 + 5(0) + D
D = 0
Therefore, the equation for displacement becomes:
s = - (3/20) t^5 + 5t
Now, to find the distance the object will move before stopping, we need to find the value of t when v = 0:
0 = - (3/4) t^4 + 5
(3/4) t^4 = 5
t^4 = (20/3)
t ≈ 1.8175 seconds
Finally, substitute the value of t into the displacement equation to find the distance traveled:
s = - (3/20) (1.8175)^5 + 5(1.8175)
s ≈ -0.8704 + 9.0875
s ≈ 8.2171 meters
Therefore, the object will move approximately 8.2171 meters before stopping.