Question

a rocket starts from rest and moves upward from the surface of the earth. for the first 10.0 s of its motion, the vertical acceleration of the rocket is given by ( a_y = (3.00 , \text{m/s}^3) , t ), where the +y-direction is upward. what is the height of the rocket above the surface of the earth at ( t = 10.0 , \text{s} )? express your answer with the appropriate units.

Explanation:

Step1: Find the velocity function by integrating acceleration

The acceleration is \( a_y = (3.00\ \text{m/s}^3)t \). To find velocity \( v_y \), we integrate \( a_y \) with respect to time \( t \). The initial velocity \( v_{0y} = 0 \) (starts from rest).

\[
v_y(t) = \int a_y\ dt + v_{0y} = \int (3.00t)\ dt + 0 = \frac{3.00}{2}t^2 + C
\]

Since \( v_{0y} = 0 \) at \( t = 0 \), \( C = 0 \), so \( v_y(t) = 1.50t^2 \).

Step2: Find the position function by integrating velocity

To find the height \( h(t) \) (position), we integrate the velocity function \( v_y(t) \) with respect to time \( t \). The initial position \( h_0 = 0 \) (starts from the surface).

\[
h(t) = \int v_y\ dt + h_0 = \int 1.50t^2\ dt + 0 = \frac{1.50}{3}t^3 + C
\]

Since \( h_0 = 0 \) at \( t = 0 \), \( C = 0 \), so \( h(t) = 0.500t^3 \).

Step3: Substitute \( t = 10.0\ \text{s} \) into the position function

Now we substitute \( t = 10.0\ \text{s} \) into \( h(t) \):

\[
h(10.0) = 0.500 \times (10.0)^3
\]

\[
h(10.0) = 0.500 \times 1000 = 500\ \text{m}
\]

Answer:

\( 500\ \text{m} \)