A space probe has two engines. Each generates the same amount of force when fired, and the directions of these forces can be independently adjusted. When the engines are fired simultaneously and each applies its force in the same direction, the probe, starting from rest, takes 52 s to travel a certain distance. How long does it take to travel the same distance, again starting from rest, if the engines are fired simultaneously and the forces that they apply to the probe are perpendicular?

If the forces are perpendicular the net force is sqrt 2 times larger than a single rocket's force. The net force and acceleration will be 1/sqrt 2 times the values when the two rockets were aligned.

Ask yourself: if X = (1/2) a t^2 is the same and a is reduced by sqrt 2, what happens to t?

15.6

To find the time it takes to travel the same distance when the engines are fired simultaneously and the forces they apply are perpendicular, we can use the concept of vector addition.

When the engines are fired in the same direction, their forces are aligned in a single vector. However, when the forces are perpendicular, we can treat them as two separate vectors, one in the x-direction and one in the y-direction.

Let's denote the time it takes to travel the distance with perpendicular forces as T. Since the probe is starting from rest, the initial velocity is zero.

When the engines are fired in the same direction, the force vector in the x-direction is Fx. Using the second law of motion (F = ma), we can write:

Fx = m*(Δvx / Δt), where m is the mass of the probe, Δvx is the change in the x-velocity, and Δt is the time interval of 52 s.

Solving for Δvx, we get:

Δvx = (Fx * Δt) / m

Since the velocity in the x-direction is changing at a constant rate, we can consider it as the average velocity during the time interval Δt. Thus, Δvx can also be written as:

Δvx = (Vx * Δt), where Vx is the average velocity in the x-direction.

Similarly, for the y-direction, when the forces are perpendicular, the force vector in the y-direction is Fy. Using the same approach, we have:

Δvy = (Fy * Δt) / m

Again, for the y-direction, Δvy can be considered as the average velocity during the time interval Δt, written as:

Δvy = (Vy * Δt), where Vy is the average velocity in the y-direction.

Now, when the forces are perpendicular, the total displacement in both x and y directions can be considered as the hypotenuse of a right triangle. This can be found using the Pythagorean theorem:

Total displacement = sqrt((Δvx)^2 + (Δvy)^2)

Since the velocity is the displacement divided by time, we have:

V = Total displacement / T

Combining the equations and rearranging, we get:

V = sqrt((Vx)^2 + (Vy)^2)

We know that Vx = Δvx/Δt and Vy = Δvy/Δt. Substituting these values, we have:

V = sqrt((Δvx)^2/Δt^2 + (Δvy)^2/Δt^2)

Simplifying the equation, we get:

V = sqrt((Δvx^2 + Δvy^2) / Δt^2)

Now, since Δvx = (Fx * Δt) / m and Δvy = (Fy * Δt) / m, we can substitute these values:

V = sqrt((Fx^2 + Fy^2) / (Δt^2 * m^2))

Since Fx = Fy (both engines generate the same force), we can simplify further:

V = sqrt((2Fx^2) / (Δt^2 * m^2))

Since the initial velocity is zero, we can find T using the equation:

T = 2V / (2Fx / m)

Substituting the value of V, we have:

T = sqrt((2Fx^2) / (Δt^2 * m^2)) * (m / Fx)

Simplifying further, we get:

T = sqrt(2) * Δt

Therefore, the time it takes to travel the same distance, starting from rest with perpendicular forces, is equal to the square root of 2 times the time interval of 52 s:

T = sqrt(2) * 52 s

T ≈ 73.57 s

So, it takes approximately 73.57 seconds to travel the same distance with the engines fired simultaneously and forces perpendicular.