Question

A rocket carrying a satellite is accelerating straight up from the earth’s surface. At 1.15 s after liftoff, the rocket clears the top of its launch platform, 63 m above the ground. After an additional 4.75 s, it is 1.00 km above the ground. Calculate the magnitude of the average velocity of the rocket for (a) the 4.75-s part of its flight and (b) the first 5.90 s of its flight.

Answer 1

Answer:

197.263157895 m/s

169.491525424 m/s

Explanation:

x Denotes position

t Denotes time

Average velocity is given by

$v_a=\dfrac{x_2-x_1}{t_2}\\\Rightarrow v_a=\dfrac{1000-63}{4.75}\\\Rightarrow v_a=197.263157895\ m/s$

The average velocity is 197.263157895 m/s

$v_a=\dfrac{x_2-x_1}{t_2}\\\Rightarrow v_a=\dfrac{1000-0}{5.9}\\\Rightarrow v_a=169.491525424\ m/s$

The average velocity is 169.491525424 m/s