On a spacecraft two engines fire for a time of 555 s. One gives the craft an acceleration in the x direction of ax = 5.10 m/s2, while the other produces an acceleration in the y direction of ay = 7.30 m/s2. At the end of the firing period, the craft has velocity components of vx = 3540 m/s and vy = 4486 m/s. Find the magnitude and direction of the initial velocity. Express the direction as an angle with respect to the +x axis.

Question

On a spacecraft two engines fire for a time of 555 s. One gives the craft an acceleration in the x direction of ax = 5.10 m/s2, while the other produces an acceleration in the y direction of ay = 7.30 m/s2. At the end of the firing period, the craft has velocity components of vx = 3540 m/s and vy = 4486 m/s. Find the magnitude and direction of the initial velocity. Express the direction as an angle with respect to the +x axis.

Answer 1

To find the magnitude and direction of the initial velocity, we can use the concept of vector addition.

1. Start by representing the given accelerations as vectors. The first engine produces an acceleration of 5.10 m/s^2 in the x direction, so its acceleration vector is (5.10, 0) m/s^2. The second engine produces an acceleration of 7.30 m/s^2 in the y direction, so its acceleration vector is (0, 7.30) m/s^2.

2. Next, determine the change in velocity in the x and y directions. The change in velocity in the x direction can be calculated using the formula Δvx = ax * Δt, where ax is the x-component acceleration and Δt is the firing time. Similarly, the change in velocity in the y direction can be calculated using the formula Δvy = ay * Δt. Substituting the given values, we get Δvx = 5.10 m/s^2 * 555 s = 2830.5 m/s and Δvy = 7.30 m/s^2 * 555 s = 4041.5 m/s.

3. Now, find the initial velocity components by subtracting the change in velocity components from the final velocity components. The initial velocity component in the x direction (vix) can be calculated as vix = vx - Δvx = 3540 m/s - 2830.5 m/s = 710.5 m/s. Similarly, the initial velocity component in the y direction (viy) can be calculated as viy = vy - Δvy = 4486 m/s - 4041.5 m/s = 445.5 m/s.

4. Using the initial velocity components (vix, viy), we can find the magnitude of the initial velocity (Vi) using the Pythagorean theorem. The magnitude can be calculated as Vi = √(vix^2 + viy^2) = √((710.5 m/s)^2 + (445.5 m/s)^2) = √(504025 m^2/s^2 + 198502.25 m^2/s^2) = √702527.25 m^2/s^2 = 838.85 m/s (rounded to two decimal places).

5. Finally, find the direction of the initial velocity (θ) with respect to the +x axis using the inverse tangent function. The direction can be calculated as θ = atan(viy/vix) = atan(445.5 m/s / 710.5 m/s) = atan(0.626) = 32.17° (rounded to two decimal places).

Therefore, the magnitude of the initial velocity is approximately 838.85 m/s and the direction is approximately 32.17° with respect to the +x axis.