Respuesta :
Answer:
a) a= [m/s²] B = [m / [tex]s^{5/2}[/tex]] 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] [tex]s^{1/2}[/tex]] = [m] / [ [tex]s^{1/2}[/tex] s²] = [m] / [ [tex]s^{2+1/2}[/tex] ]
B = [m / [tex]s^{5/2}[/tex]]
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
