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) a= [m/s²]     B = [m / ]     b) v = A t - 2/3 B √t³  

Explanation:

The equation gives us for acceleration is a (t) = A - B √t  

a) The unit of constants

Since the function is the acceleration and you have units of [m / s²] the quantities the right must have the same units, therefore, the counter A is in [m / s²] and the other term is B for the time

   B t ½ = [m / s²]

   B [s ½] = [m / s²]

   B = [m / s²] ] = [m] / [  s²] = [m] / [  ]

   B = [m / ]

b) To calculate the rocket speed we must use the definition of acceleration and integrate

    a = dv / dt

    dv = a dt

    ∫ dv = ∫ (A - B √t) dt

    v = A (t) - B  √t³   / 3/2

    v = At - 2/3 B  √t³

We evaluated entered the initial point where the zero velocity equals a zero time and the final point with velocity v and time t

     v = A t - 2/3 B √t³  

c) To find the position we repeat the same proceeding for speed

     v = dx / dt

     dx = v dt

     dx = / v dt

     ∫ dx = ∫ (A t - 2/3 B √t³ ) dt

     x =  At² / 2 - 2/3 B √t⁵  / 5/2

     x = ½ A t² - 2/3 2/5 B √t⁵

Evaluating from the start point with zero position for a time of zero or to the end point with position x at time t

      x = ½ A t² - 4/15 B √t⁵

unbalanced force

explanation: