In this example we will use the following equation

aav,x=v2x−v1xt2−t1=ΔvxΔt

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)v1x(b)v1x(c)v1x(d)v1x====0.8m/s,1.6m/s,−0.4m/s,−1.6m/s,v2xv2xv2xv2x====1.2m/s(speedingup);1.2m/s(slowingdown);−1.0m/s(speedingup);−0.8m/s(slowingdown).

If t1=2s
and t2=4s
in each case, find the average acceleration for each set of data. For a fifth set of maneuvers, the astronaut’s partner calculates her average acceleration over the 2  s
interval to be -0.7  m/s2
. If the initial velocity of the astronaut was -0.4  m/s
, what was her final velocity?

### Average Acceleration Calculation:

The formula for average acceleration is:

\[
a_{\text{av,x}} = \frac{v_{2x} - v_{1x}}{t_2 - t_1} = \frac{\Delta v_x}{\Delta t}
\]

Where:
- \(v_{1x}\) is the initial velocity
- \(v_{2x}\) is the final velocity
- \(t_1\) and \(t_2\) are the initial and final times respectively

Given that \( t_1 = 2 \, \text{s} \) and \( t_2 = 4 \, \text{s} \), the time interval \( \Delta t = t_2 - t_1 = 2 \, \text{s} \).

### Case-by-Case Calculation:

#### (a) \( v_{1x} = 0.8 \, \text{m/s}, v_{2x} = 1.2 \, \text{m/s} \) (speeding up)

\[
a_{\text{av,x}} = \frac{1.2 \, \text{m/s} - 0.8 \, \text{m/s}}{2 \, \text{s}} = \frac{0.4 \, \text{m/s}}{2 \, \text{s}} = 0.2 \, \text{m/s}^2
\]

#### (b) \( v_{1x} = 1.6 \, \text{m/s}, v_{2x} = 1.2 \, \text{m/s} \) (slowing down)

\[
a_{\text{av,x}} = \frac{1.2 \, \text{m/s} - 1.6 \, \text{m/s}}{2 \, \text{s}} = \frac{-0.4 \, \text{m/s}}{2 \, \text{s}} = -0.2 \, \text{m/s}^2
\]

#### (c) \( v_{1x} = -0.4 \, \text{m/s}, v_{2x} = -1.0 \, \text{m/s} \) (speeding up)

\[
a_{\text{av,x}} = \frac{-1.0 \, \text{m/s} - (-0.4 \, \text{m/s})}{2 \, \text{s}} = \frac{-0.6 \, \text{m/s}}{2 \, \text{s}} = -0.3 \, \text{m/s}^2
\]

#### (d) \( v_{1x} = -1.6 \, \text{m/s}, v_{2x} = -0.8 \, \text{m/s} \) (slowing down)

\[
a_{\text{av,x}} = \frac{-0.8 \, \text{m/s} - (-1.6 \, \text{m/s})}{2 \, \text{s}} = \frac{0.8 \, \text{m/s}}{2 \, \text{s}} = 0.4 \, \text{m/s}^2
\]

### Fifth Set of Data:

For this set, we are given:
- Average acceleration \( a_{\text{av,x}} = -0.7 \, \text{m/s}^2 \)
- Initial velocity \( v_{1x} = -0.4 \, \text{m/s} \)
- Time interval \( \Delta t = 2 \, \text{s} \)

We can use the average acceleration formula to solve for the final velocity \( v_{2x} \):

\[
a_{\text{av,x}} = \frac{v_{2x} - v_{1x}}{\Delta t}
\]

Substituting the values:

\[
-0.7 = \frac{v_{2x} - (-0.4)}{2}
\]

\[
-0.7 \times 2 = v_{2x} + 0.4
\]

\[
-1.4 = v_{2x} + 0.4
\]

\[
v_{2x} = -1.8 \, \text{m/s}
\]

### Summary of Results:

1. **Case (a)**: \( a_{\text{av,x}} = 0.2 \, \text{m/s}^2 \)
2. **Case (b)**: \( a_{\text{av,x}} = -0.2 \, \text{m/s}^2 \)
3. **Case (c)**: \( a_{\text{av,x}} = -0.3 \, \text{m/s}^2 \)
4. **Case (d)**: \( a_{\text{av,x}} = 0.4 \, \text{m/s}^2 \)
5. **Fifth set**: Final velocity \( v_{2x} = -1.8 \, \text{m/s} \)

Would you like further clarification on any of these steps or more details?

---

**Further Questions:**

1. What is the significance of positive vs negative acceleration?
2. How does changing the time interval affect the average acceleration?
3. How can we interpret negative velocities in these calculations?
4. What happens if the final velocity is zero in any of the above cases?
5. Can acceleration remain constant while velocity changes?

**Tip**: Remember that acceleration can be negative (deceleration) when an object slows down or moves in the opposite direction of its initial velocity.

In this example we will use the following equation aav,x=v2x−v1xt2−t1=ΔvxΔt 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)v1x====0.8m/s, v2x====1.2m/s (speeding up); (b)v1x====1.6m/s, v2x====1.2m/s (slowing down); (c)v1x====-0.4m/s, v2x====-1.0m/s (speeding up); (d)v1x====-1.6m/s, v2x====-0.8m/s (slowing down). If t1=2s and t2=4s in each case, find the average acceleration for each set of data. For a fifth set of maneuvers, the astronaut’s partner calculates her average acceleration over the 2 s interval to be -0.7 m/s2. If the initial velocity of the astronaut was -0.4 m/s, what was her final velocity?

This problem explains how to calculate the average acceleration of an astronaut during a spacewalk using velocity measurements. We examine various cases of speeding up and slowing down, and solve for final velocity given the initial conditions and time intervals. Suitable for Grades 10-12 students studying kinematics.

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