Question

Two rockets are flying in the same direction and are side by side at the instant their retrorockets fire. Rocket A has an initial velocity of +4600 m/s, while rocket B has an initial velocity of +8200 m/s. After a time t both rockets are again side by side, the displacement of each being zero. The acceleration of rocket A is -18 m/s2. What is the acceleration of rocket B?

Answer 1

Answer:

$a_2\ =\ -33.65\ m/s^2$

Explanation:

Given,

For the first rocket,

• Initial velocity of the first rocket A = $u_1\ =\ 4600\ m/s.$

• Acceleration of the first rocket = $a_1\ =\ -18\ m/s^2$

For the second rocket,

• Initial velocity of the second rocket B = $u_2\ =\ 8200 m/s.$
• Displacement of both the rockets A and B = s = 0 m

Fro the first rocket,

Let 't' be the time taken by the first rocket A for whole the displacement

$\therefore s\ =\ u_1t\ +\ \dfrac{1}{2}a_1t^2\\\Rightarrow 0\ =\ 4600t\ -\ 0.5\times 18t^2\\\Rightarrow t\ =\ \dfrac{4600}{0.5\times 18}\\\Rightarrow t\ =\ 511.11 sec$

Let $a_2$ be the acceleration of the second rocket B for the same time interval

from the kinematics,

$\therefore s\ =\ ut\ +\ \dfrac{1}{2}at^2\\\Rightarrow s\ =\ u_2t\ +\ \dfrac{1}{2}a_2t^2\\\Rightarrow a_2\ =\ \dfrac{2s\ -\ 2u_2t}{t^2}\\\Rightarrow a_2\ =\ \dfrac{0\ -\ 2u_2t}{t^2}$

$\Rightarrow a_2\ =\ -\dfrac{2u_2}{t}\\\Rightarrow a_2\ =\ -\dfrac{2\times 8600}{511.11}\\\Rightarrow a_2\ =\ -33.65\ m/s^2$

Hence the acceleration of the second rocket B is -33.65\ m/s^2.