A spacecraft is travelling with a velocity of v_{0 x} = 4570 m/s along the +x direction. Two...

Question:

A spacecraft is travelling with a velocity of {eq}\rm v_{0 x} = 4570\ m/s {/eq} along the +x direction. Two engines are turned on for a time of {eq}\rm 873\ s {/eq}. One engine gives the spacecraft an acceleration in the {eq}+x {/eq} direction of {eq}\rm a_x = 1.20\ m/s^2 {/eq}, while the other gives it an acceleration in the {eq}+y {/eq} direction of {eq}\rm a_y = 7.70\ m/s^2 {/eq}. At the end of the firing, find the following.

(a) {eq}v_x {/eq}.

(b) {eq}v_y {/eq}.

Velocity:

The velocity represents how briskly the displacement of the body with the change in time. The velocity of an object determined if the displacement of the object and the time is given. In a two dimensional plane, the velocity has two components: horizontal component and vertical velocity.

Answer and Explanation:


Given Data

  • The initial velocity of the spacecraft in the horizontal direction is {eq}{v_{0x}} = 4570\;{\rm{m/sec}} {/eq}.
  • The time is {eq}t = 873\;\sec {/eq}.
  • The acceleration in the horizontal direction is {eq}{a_x} = 1.20\;{\rm{m/se}}{{\rm{c}}^{\rm{2}}} {/eq}.
  • The acceleration in the vertical direction is {eq}{a_y} = 7.70\;\;m/se{c^2} {/eq}.


(a)

The expression for the horizontal direction is,

{eq}{v_x} = {v_{0x}} + {a_x}t {/eq}


Substitute the given values in the above expression.

{eq}\begin{align*} {v_x} &= 4570\;{\rm{m/sec}} + 1.20\;{\rm{m/se}}{{\rm{c}}^{\rm{2}}} \times 873\;\sec \\ &= 4570\;{\rm{m/sec}} + 1047.6\;{\rm{m/sec}}\\ &= 5617.6\;{\rm{m/sec}} \end{align*} {/eq}


Thus, the horizontal velocity is {eq}5617.6\;{\rm{m/sec}} {/eq}.


(b)


The expression for the horizontal direction is,

{eq}{v_y} = {v_{0y}} + {a_y}t {/eq}


Substitute the given values in the above expression.

{eq}\begin{align*} {v_y} &= 0\;{\rm{m/sec}} + 7.70\;{\rm{m/se}}{{\rm{c}}^{\rm{2}}} \times 873\;\sec \\ &= 0\;{\rm{m/sec}} + 6722.1\;{\rm{m/sec}}\\ &= 6722.1\;{\rm{m/sec}} \end{align*} {/eq}


Thus, the vertical velocity is {eq}6722.1\;{\rm{m/sec}} {/eq}.


Learn more about this topic:

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Linear Velocity: Definition & Formula

from MCAT Prep: Help and Review

Chapter 14 / Lesson 12
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