Question

A spacecraft on its way to Mars has small rocket engines mounted on its hull; one on its left surface and one on its back surface. At a certain time, both engines turn on. The one on the left gives the spacecraft an acceleration component in the x direction of

ax = 5.10 m/s2,
while the one on the back gives an acceleration component in the y direction of
ay = 7.30 m/s2.
The engines turn off after firing for 670 s, at which point the spacecraft has velocity components of
vx = 3670 m/s and vy = 4378 m/s.
What was the magnitude and the direction of the spacecraft's initial velocity before the engines were turned on? Express the magnitude as m/s and the direction as an angle measured counterclockwise from the +x axis.

magnitude m/s
direction ° counterclockwise from the +x-axis

Answer 1

a) v = 517.99 m / s, b) θ = 296.3º

Step-by-step explanation:

This is an exercise in kinematics, we are going to solve each axis independently

X axis

the acceleration is aₓ = 5.10 1 / S², they are on for t = 670 s and reaches a speed of vₓ= 3670 m / s, let's use the relation

vₓ = v₀ₓ + aₓ t

v₀ₓ = vₓ - aₓ t

v₀ₓ = 3670 - 5.10 670

v₀ₓ = 253 m / s

Y axis

the acceleration is ay = 7.30 m / s², with a velocity of 4378 m / s after

t = 670 s

v_y = v_{oy} + a_y t

v_{oy} = v_y - a_y t

v_oy} = 4378 - 7.30 670

v_{oy} = -513 m / s

to find the velocity modulus we use the Pythagorean theorem

v =
`√(v_o_x^2 + v_o_y^2)`

v =
`√(253^2 +513^2)`

v = 517.99 m / s

to find the direction we use trigonometry

tan θ ’=
`(v_o_y)/(v_o_x)`

θ'= tan⁻¹
`(voy)/(voy)`

θ'= tan⁻¹ (-513/253)

tea '= -63.7

the negative sign indicates that it is below the ax axis, in the fourth quadrant

to give this angle from the positive side of the axis ax

θ = 360 - θ

θ = 360 - 63.7

θ = 296.3º