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 velicity of +5800m/s, while Rocket B has an initial velocity of +8600m/s. After a time t both rockets are again side by side, the displacement of each being zero. The acceleration of rocket A is -15m/s^2. What is the acceleration of rocket B?

Answer 1

Answer:

-22.2 m/s²

Explanation:

The equation for position x for a constant acceleration a, time t and initial velocity v₀, initial position x₀:

(1) $x=\frac{1}{2}at^2+v_0t+x_0$

For rocket A the initial and final position: x = x₀= 0. Using these values in equation 1 gives:

(2) $0=\frac{1}{2}at^2+v_0t$

Solving for time t:

$-\frac{1}{2}at^2=v_0t$

(3) $t=-\frac{2v_0}{a}$

The times for both rockets must be equal, since they start and end at the same location. Using equation 3 for rocket A and B gives:

(4) $\frac{v_{0A}}{a_A}=\frac{v_{0B}}{a_B}$

Solving equation 4 for acceleration of rocket B:

(5) $a_B=a_A\frac{v_{0B}}{v_{0A}}$