Question

Between t = 0 and t = t0, a rocket moves straight upward with an acceleration given by a(t)=A−Bt1/2a(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?

Answer 1

Answer:

a)  A = [m /s²] ,  B = [m / $t^{3/2}$] , b) v = a to + 2B/3 $t^{3/2}$

Explanation:

a) to define the units of the constants, we see that the left part has units of acceleration so the right part must also have the same units

     A = [m /s²]

     B   = [m / s²]

     B = [m / $t^{3/2}$]

b) We must use the definition of acceleration

      a = dv / dt

      dv = adt

We integrate

       

    ∫ dv = ∫ (A + B $t^{1/2}$) dt

    v = a t + B 2/3 $t^{3/2}$

We evaluate between:

the lower limit t = 0 v = o and the upper limit t = to v = v

    v = a to + 2B/3 $t^{3/2}$

c) Let's use the definition of speed

    v = dx / dt

    dx = v dt

    ∫ dx = ∫ (a t + 2B/3 $t^{3/2}$) dt

    x = a t²/2 + 2b/3 2/5 $t^{5/2}$

For the interval t = 0 and t = to, we evaluate the integral

    x = A/2 t₀² + 4B/15 √t₀⁵