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. part b what is the speed of the rocket when it is 245 m above the surface of the earth? express your answer with the appropriate units.

Explanation:

Step1: Find the velocity function from acceleration

The acceleration is \( a_y = 3.00t \, \text{m/s}^3 \). Velocity is the integral of acceleration with respect to time. Since the rocket starts from rest, the initial velocity \( v_{0y} = 0 \).
\[
v_y(t)=\int a_y dt=\int 3.00t dt = 1.50t^2 + C
\]
Using the initial condition \( t = 0, v_y = 0 \), we find \( C = 0 \), so \( v_y(t)=1.50t^2 \).

Step2: Find the position function from velocity

Position is the integral of velocity with respect to time. The initial position \( y_0 = 0 \).
\[
y(t)=\int v_y dt=\int 1.50t^2 dt = 0.50t^3 + D
\]
Using the initial condition \( t = 0, y = 0 \), we find \( D = 0 \), so \( y(t)=0.50t^3 \).

Step3: Find time when \( y = 245 \, \text{m} \)

Set \( y(t)=245 \):
\[
0.50t^3 = 245
\]
\[
t^3=\frac{245}{0.50}=490
\]
\[
t=\sqrt[3]{490}\approx7.88 \, \text{s}
\]
(We check that \( t = 7.88 \, \text{s} < 10.0 \, \text{s} \), so the acceleration function still applies.)

Step4: Find velocity at \( t \approx 7.88 \, \text{s} \)

Use \( v_y(t)=1.50t^2 \):
\[
v_y=1.50\times(7.88)^2
\]
\[
v_y=1.50\times62.0944\approx93.1 \, \text{m/s}
\]

Answer:

The speed of the rocket is approximately \(\boldsymbol{93.1 \, \text{m/s}}\).