Answer:
a) The velocity of the object as a function of time, v(t) is 2·b·t
b) The acceleration of the function of time, a(t) is 2·b
c) The time at which the object is at the flagpole is t = √(c/b)
Step-by-step explanation:
The function that gives the position of the object north of the flagpole, x(t) is presented as follows;
x(t) = b·t² - c (b and c are constants)
a) The velocity of the object as a function of time, v(t), is derived as follows
v(t) = x'(t) = d(b·t² - c)/dt = 2·b·t
The velocity of the object as a function of time, v(t) = 2·b·t
b) The acceleration of the function of time, a(t) = v'(t) = d(2·b·t)/dt = 2·b
c) The time at which the object is at the flagpole is given by the x-intercept of the function, where x(t) = 0, as follows;
At the x-intercept, we have, x(t) = 0 and x(t) = b·t² - c
∴ 0 = b·t² - c, which gives
b·t² = c
t² = c/b
t = ±√(c/b), we reject the negative value to get;
The time at which the object is at the flagpole, t = √(c/b).