91Ó°ÊÓ

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.

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.

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

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

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-

$$\begin{gathered} a=2×\frac{63\mathrm{m}−0}{(1.15\mathrm{s}{)}^2} \\ ≈95.27\frac{\mathrm{m}}{{\mathrm{s}}^2} \end{gathered}$$

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

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

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-

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

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-

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

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-

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

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