Q59P - University Physics with Modern Physics

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 5.90 s first  of its flight.

Step-by-Step Solution

Answer

a) magnitude of the average velocity for $t=4.75 s$ is $226.75\hspace{0.33em}\frac{m}{s}$ upwards and b) magnitude of the average velocity for t =4.75 s is $281.0465\hspace{0.33em}\frac{m}{s}$ upwards.

Step 1: Identification of the given data

The given data can be expressed below as

• The rocket has lifted off after 1.15 s.
• The rocker has lifted of 63 m above the ground.
• The rocket was above the ground after 4.75 s.

Step 2: Significance of the Newton’s first law for the rocket

This law states that an object will continue to be in uniform motion unless an external force acts upon the particle.

The equation of the displacement of the rocket gives the acceleration of the rocket and the equation of the velocity gives the average velocity of the rocket at different time intervals.

Step 3: Determination of the magnitude of the average velocity of the rocket

From Newton’s first law, the displacement of the rocket can be expressed as:

$$\begin{gathered} d=v_{i}t+\frac{1}{2}a{t}^2 \\ a=2\times \frac{d-v_{i}t}{t^2} \end{gathered}$$

As the rocket was at rest, the initial velocity of the rocket $v_i$ is zero, the distance d for the rocket is 63m and the time in which the rocket has lifted off is 1.15 s.

Substituting the values in the above equation, we get-

$$\begin{gathered} a=2\times \frac{63\mathrm{m}-0}{(1.15\mathrm{s}{)}^2} \\ \approx 95.27\frac{\mathrm{m}}{{\mathrm{s}}^2} \end{gathered}$$

a) the final velocity of the rocket at t=4.75 s is-

$$\begin{gathered} v_f=v_i+at \\ =0+\left(95.27\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\times \left(4.75\mathrm{s}\right) \\ \approx 452.55\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Hence, the average velocity of the rocket for t=4.75 s is-

$v=\frac{V_i+v_f}{2}$

Substituting the values in the above equation, we get-

$$\begin{gathered} v=\frac{0+452.55\frac{m}{s}}{2} \\ \approx 226.75\frac{m}{s} \end{gathered}$$

Thus, magnitude of the average velocity for t =4.75 s is  $226.75\frac{m}{s}$ upwards.

b) the final velocity of the rocket at t =5.90 s is-

$$\begin{gathered} v_f=v_i+at \\ =0+\left(95.27\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\times \left(5.90\mathrm{s}\right) \\ \approx 562.093\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Hence, the average velocity of the rocket for is-

$v=\frac{v_i+v_f}{2}$

Substituting the values in the above equation, we get-

$$\begin{gathered} v=\frac{0+562.093\frac{\mathrm{m}}{\mathrm{s}}}{2} \\ \approx 281.0465\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Thus, magnitude of the average velocity for t=4.75 s is $281.0465\frac{m}{s}$ upwards .