Marble Velocity help URGENT!

Question

A student determines that the y - component of the acceleration of a marble being pushed by an unknown force is a y (t) = - 3t , where +y is the upward direction . The marble starts at a height of 2 m and has an initial velocity of 10 m /s (upward).

(a) Use the relationships between position, velocity, and acceleration to find the formula vy(t). Then find the velocity at 2 s.

(b) Use the relationships between position, velocity, and acceleration to find the formula y(t). Then find the height at 2 s.

(c) Use the equations determined in the previous sections to find the maximum height of the marble.

Answer 1

The notation isn't correct, but it seems you mean the acceleration function is a_y(t) = –jt where j = 3m/s³ will ensure expressions have the right units.

Integrate this to get v_y(t) – v_y(0) = ∫₀^ta_y(u)du = –jt²/2.

Integrate this to get y(t) – y(0) = ∫₀^tv_y(u)du = v_y(0)t – jt³/6.

The time of the maximum height is the root t → T given by

0 = v_y(T) = v_y(0) – jT²/2.