A particle moves along a straight path with an acceleration that varies with time according to the equation a=2t. The object is initially at rest at the origin. Determine the object's displacement at time t= 2 seconds.
a(t) = 2t
v(t) = t^2 since v(0) = 0
x(t) = 1/3 t^3 since x(0) = 0
Now find x(2)
d^2 x/dt^2 = 2 t
dx/dt = t^2 + 0
x = t^3 / 3 + 0 + 0
when t= 2
x = 8/3
To determine the object's displacement at time t = 2 seconds, we need to integrate the equation for acceleration with respect to time and substitute the limits of integration.
Given that the acceleration equation is a = 2t, we can integrate it to find the velocity equation v(t) using the formula:
v(t) = ∫(a)dt
= ∫(2t)dt
Integrating 2t with respect to t gives us:
v(t) = t^2 + C
Next, we can find the constant of integration, C, by considering that the object starts at rest. Since the object is initially at rest at the origin, its initial velocity (v₀) is 0 when t = 0. Substituting these values into the equation, we get:
0 = (0)^2 + C
C = 0
Thus, the velocity equation becomes:
v(t) = t^2
Finally, to find the displacement at time t = 2 seconds, we need to integrate the velocity equation with respect to time:
s(t) = ∫(v(t))dt
= ∫(t^2)dt
Integrating t^2 with respect to t gives us:
s(t) = (1/3)t^3 + D
Now, we can find the constant of integration, D, by considering the initial displacement of the object. Since the object starts at the origin, its initial displacement (s₀) is 0 when t = 0. Substituting these values into the equation, we get:
0 = (1/3)(0)^3 + D
D = 0
Therefore, the displacement equation becomes:
s(t) = (1/3)t^3
Finally, substituting t = 2 seconds into the displacement equation, we can calculate the object's displacement at that time:
s(2) = (1/3)(2)^3
= (1/3)(8)
= 8/3 units
So, the object's displacement at t = 2 seconds is 8/3 units.