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Q. 2.3

Aceleration in a space walk

In this example we will use Equation 2.3 to calculate the acceleration of an astronaut on a space walk. The astronaut has left the space shuttle on a tether to test a new personal maneuvering device. She moves along a straight line directly away from the shuttle. Her onboard partner measures her velocity before and after certain maneuvers, and obtains these results:

$a_{\mathrm{av},x}=\frac{\upsilon_{2x}-\upsilon_{1x}}{t_2-t_1}=\frac{\Delta \upsilon_x}{\Delta t}.$                        (2.3)

(a) $\upsilon _{1x} = 0.8 m/s, \upsilon_{2x} = 1.2 m/s$ (speeding up);

(b) $\upsilon _{1x} = 1.6 m/s, \upsilon _{2x} = 1.2 m/s$ (slowing down);

(c) $\upsilon _{1x} = -0.4 m/s, \upsilon _{2x} = -1.0 m/s$ (speeding up);

(d) $\upsilon _{1x} = -1.6 m/s, \upsilon _{2x} = -0.8 m/s$ (slowing down).

If $t_1 = 2 s and t_2 = 4 s$ in each case, find the average acceleration for each set of data.

Verified Solution

SET UP We use the diagram in Figure 2.11 to organize our data.

SOLVE To find the astronaut’s average acceleration in each case, we use
the definition of average acceleration (Equation 2.3): $a_\mathrm{{av, x}} = ∆\upsilon _x/∆t.$
The time interval is ∆t = 2.0 s in all cases; the change in velocity in each case is $\Delta \upsilon_x = \upsilon _{2x} – \upsilon_{1x}.$

Part (a): $a_{\mathrm{av,x}}=\frac{1.2 m/s-0.8 m/s}{4s-2s}=+0.2 m/s^2 ;$

Part (b): $a_{\mathrm{av,x}}=\frac{1.2 m/s-0.8 m/s}{4s-2s}=-0.2 m/s^2 ;$

Part (c):$a_{\mathrm{av,x}}=\frac{-1.0 m/s-(-0.4 m/s)}{4s-2s}=-0.3 m/s^2 ;$

Part (d): $a_{\mathrm{av,x}}=\frac{-0.8 m/s-(-1.6 m/s)}{4s-2s}=+0.4 m/s^2.$

REFLECT The astronaut speeds up in cases (a) and (c) and slows down in (b) and (d), but the average acceleration is positive in (a) and (d) and negative in (b) and (c). So negative acceleration does not necessarily indicate a slowing down.

Practice Problem: For a fifth set of maneuvers, the astronaut’s partner calculates her average acceleration over the 2 s interval to be -0.5 m/s².
If the initial velocity of the astronaut was -0.4 m/s, what was her final velocity? Did she speed up or slow down over the 2 s interval? Answers: -1.4 m/s, speed up.