On a spacecraft two engines fire for a time of 615 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 = 3805 m/s and vy = 4148 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 615 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 = 3805 m/s and vy = 4148 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 of the spacecraft, we need to use the given information about the acceleration and firing time.

First, let's calculate the change in velocity for each component separately. We can use the following equation of motion:

Δv = a * t

Where Δv is the change in velocity, a is the acceleration, and t is the time.

For the x-direction:
Δvx = ax * t = 5.10 m/s² * 615 s = 3136.5 m/s

For the y-direction:
Δvy = ay * t = 7.30 m/s² * 615 s = 4489.5 m/s

Next, let's find the initial velocity in each component by subtracting the change in velocity from the final velocity:

Initial velocity in the x-direction:
vix = vx - Δvx = 3805 m/s - 3136.5 m/s = 668.5 m/s

Initial velocity in the y-direction:
viy = vy - Δvy = 4148 m/s - 4489.5 m/s = -341.5 m/s

Now we have the initial velocity components (vix, viy) in the x and y directions. To find the magnitude of the initial velocity (vi) using the Pythagorean theorem:

vi = √(vix² + viy²)
= √((668.5 m/s)² + (-341.5 m/s)²)
≈ √(446.22 m² + 116.62 m²)
≈ √(199.39 m²)
≈ 14.11 m/s

So, the magnitude of the initial velocity is approximately 14.11 m/s.

To find the direction of the initial velocity, we can use the arctangent function:

θ = tan^(-1)(viy / vix)
= tan^(-1)(-341.5 m/s / 668.5 m/s)

Using a calculator or math software, we find that:
θ ≈ -27.55°

Since the question asks for the direction with respect to the +x axis, we need to convert this to a positive angle.

The angle with respect to the +x axis is:
θ = 180° - |-27.55°| ≈ 180° - 27.55° ≈ 152.45°

Therefore, the magnitude and direction of the initial velocity of the spacecraft are approximately 14.11 m/s at an angle of 152.45° with respect to the +x axis.