A rocket starts from rest and moves upward from the surface of the earth. For the first 10s of its motion, the vertical acceleration of the rocket is given by a_y =( 2.70m/s^3)t, where the +y-direction is upward.
1) What is the height of the rocket above the surface of the earth at t = 10 s?
2) What is the speed of the rocket when it is 345 m above the surface of the earth?
what is the speed of the rocket when it is 345 m above the surface of the earth?
140 m
To find the height of the rocket above the surface of the earth at t = 10 s, you need to integrate the equation for acceleration with respect to time twice to get the position equation.
1) Integrating the acceleration equation once will give you the velocity equation:
v_y = ∫ (a_y) dt
v_y = ∫ (2.70m/s^3)t dt
v_y = (2.70/2)s^2 t^2 + C1
Since the rocket starts from rest, the initial velocity is zero (v_y = 0) when t = 0, so C1 = 0.
v_y = (2.70/2)s^2 t^2
2) Integrating the velocity equation again will give you the position equation:
y = ∫ (v_y) dt
y = ∫ [(2.70/2)s^2 t^2] dt
y = (2.70/6)s^2 t^3 + C2
To find C2, you need a boundary condition. The height of the rocket above the surface of the earth at t = 0 is zero (y = 0), so C2 = 0.
y = (2.70/6)s^2 t^3
Now, substitute t = 10 s into the position equation to find the height of the rocket above the surface of the earth at t = 10 s:
y = (2.70/6)s^2 (10^3)
y = 45 * 10^3
y = 45000 m
So, the height of the rocket above the surface of the earth at t = 10 s is 45000 meters.
To find the speed of the rocket when it is 345 m above the surface of the earth, you can use the velocity equation.
1) Set y = 345 m in the position equation:
345 = (2.70/6)s^2 t^3
2) Solve for t:
t = (345 * 6/(2.70 * s^2))^(1/3)
3) Substitute the value of t back into the velocity equation to find the speed:
v_y = (2.70/2)s^2 t^2
Remember to convert the units of the answer to the desired unit (e.g., kilometers per hour or meters per second).