The spacecraft shown in Figure 2.14a is traveling with a velocity of +3250 m/s. Suddenly the retrorockets are fired, and the spacecraft begins to slow down with an acceleration whose magnitude is 10.0 m/s2. What is the velocity of the spacecraft when the displacement of the craft is +215 km, relative to the point where the retrorockets began firing?
 |
|
Reasoning
Since the spacecraft is slowing down, the acceleration must be opposite to the velocity. The velocity points to the right in the drawing, so the acceleration points to the left, in the negative direction; thus, a
=–10.0 m/s2. The three known variables are listed as follows:
Spacecraft Data
x
a
v
v0
t
+215 000 m
–10.0 m/s2
?
+3250 m/s
The final velocity v of the spacecraft can be calculated using Equation 2.9, since it contains the four pertinent variables.
Solution
From Equation 2.9 (
), we find that
 | |
=+215 km), but each arises in a different part of the motion. The answer v=+2500 m/s corresponds to the situation in Figure 2.14a, where the spacecraft has slowed to a speed of 2500 m/s, but is still traveling to the right. The answer v=–2500 m/s arises because the retrorockets eventually bring the spacecraft to a momentary halt and cause it to reverse its direction. Then it moves to the left, and its speed increases due to the continually firing rockets. After a time, the velocity of the craft becomes v=–2500 m/s, giving rise to the situation in Figure 2.14b. In both parts of the drawing the spacecraft has the same displacement, but a greater travel time is required in part b compared to part a.
