Question

An object’s velocity is measured to be vx1t2 = a - bt 2 , where a = 4.00 m>s and b = 2.00 m>s 3 . at t = 0 the object is at x = 0. (a) calculate the object’s position and acceleration as functions of time. (b) what is the object’s maximum positive displacement from the origin?

Answer 1

The object's position and acceleration can be found by integrating and differentiating the velocity function. The maximum positive displacement from the origin occurs when the object's position is zero.

Step-by-step explanation:

(a) To find the object's position as a function of time, we need to integrate the velocity function with respect to time. Integrate vx1(t^2) = a - bt^2 to find x(t):

x(t) = ∫ (a - bt^2) dt = ax - ⅓bt^3 + C

Since at t = 0 the object is at x = 0, we have the initial condition ax = C. Therefore, the object's position is given by x(t) = ax - ⅓bt^3 + ax = 2ax - ⅓bt^3.

To find the object's acceleration as a function of time, we need to take the derivative of the velocity function with respect to time. Differentiate vx1(t^2) = a - bt^2 to find a(t):

a(t) = d(vx1(t^2))/dt = -2bt

(b) To find the object's maximum positive displacement from the origin, we need to find the maximum value of x(t) when t > 0. Since the object is at x = 0 when t = 0, the maximum positive displacement occurs when x(t) = 0, which gives:

2ax - ⅓bt^3 = 0

2ax = ⅓bt^3

t = (⅓b/a)^(1/3)