Question

Between t = 0 and t = t0, a rocket moves straight upward with an acceleration given by a(t)=A−Bt1/2, where A and B are constants. (a) If x is in meters and t is in seconds, what are the units of A and B? (b) If the rocket starts from rest, how does the velocity vary between t = 0 and t = t0? (c) If its initial position is zero, what is the rocket’s position as a function of time during this same time interval?

a) Option a: A in m/s^2 and B in s^(1/2)/m^(1/2); Option b: Linear increase; Option c: x(t) = At - (2B/3)t^(3/2)
b) Option a: A in m/s^2 and B in s^(1/2)/m^(1/2); Option b: Decreases linearly; Option c: x(t) = At - (2B/3)t^(3/2)
c) Option a: A in m/s^2 and B in s/m^(1/2); Option b: Decreases with t; Option c: x(t) = At^2 - 2Bt^(3/2)
d) Option a: A in m/s^2 and B in s/m; Option b: Linear increase; Option c: x(t) = At^2 - 2Bt^(3/2)

Answer 1

The units of A are m/s^2 and the units of B are s^(1/2)/m^(1/2). The velocity of the rocket increases linearly as time increases. The position of the rocket as a function of time can be determined using the position function x(t) = At - (2B/3)t^(3/2).

Step-by-step explanation:

To determine the units of A and B in the equation a(t) = A - Bt^1/2, we need to consider the units of acceleration and time. The unit of acceleration is meters per second squared (m/s^2) and the unit of time is seconds (s). Therefore, the units of A must be in m/s^2 and the units of B must be in s^(1/2)/m^(1/2).

Since the rocket starts from rest, its initial velocity is zero. As time increases from t = 0 to t = t0, the velocity of the rocket will increase linearly. This means that the velocity will steadily increase over time.

If the rocket's initial position is zero, its position as a function of time during this time interval can be determined by integrating the acceleration function. The position function is given by x(t) = At - (2B/3)t^(3/2).