Arocket is fired at an angle from the top of a tower of height h 0 = 50.0 m. because of the design of the engines, its position coordinates are of the form x1t2 = a + bt 2 and y1t2 = c + dt 3 , where a, b, c, and d are constants. the acceleration of s the rocket 1.00 s after firing is a = 14.00d n + 3.00e n 2 m> s 2 . take the origin of coordinates to be at the base of the tower. (a) find the constants a, b, c, and d, including their si units. (b) at the instant after the rocket is fired, what are its acceleration vector and its velocity? (c) what are the x- and y-components of the rocket’s velocity 10.0 s after it is fired, and how fast is it moving? (d) what is the position vector of the rocket 10.0 s after it is fired?

a)   A = 0 [m] , B = 7.00 m/s² , C = 50.0 [m] , D = 0.500 m/s³

Explanation:

This is an exercise in kinematics in two dimensions, the easiest way to solve it is to work each axis independently

a) Value of the constants

In the expression the left side has units of bone position meters [m], so the right side also has to have the same units

       A = 0 [m]

       B = [m / s2]

       C = 50.0 [m]

       D = [m / s3]

To find the other two constants we must know the acceleration formula, let's use its definition

       a = dv / dt = d²v / dt²

      vₓ = dx / dt = 2Bt

      d²x / dt² = 2B

      aₓ = 2B

      = dy / dt = 3D t2

      d²y / dt² = 6D t

They give us the acceleration for time t = 1.00s

      aₓ = 2B = 14.00 m / s2

      B = 7.00 m / s²

      = 6D t = 3.00

      D = 3.00 / 6 1

      D = 0.500 m / s³

b) The acceleration vector is

       a = aₓ i ^ + j ^

       a = 14 i ^ + 3 t j ^

       v = vx i ^ + vyj ^

       v = 14 t i ^ + 1.5 t²

c)  velocity for t= 10.0 s

          v = 14 t i ^ + 1.5 t²

          v = 14 10 i^ + 1.5 10² j^

v= (140 i^ + 150 j^ ) m/s

d)  Position for t = 10.0 s

        x = 7 t² = 7 10²

        x = 700 m

        y = 50.0 +0.5 t³ = 50.0 + 0.5 10³

        y = 550 m

        D = (700 i ^ + 550 j ^) m

the answer is b

explanation:

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