Chapter 8: Problem 28
Particle acceleration If a particle moves along a coordinate line with a constant acceleration \(a\) (in \(\mathrm{cm} / \mathrm{sec}^{2}\) ), then at time \(t\) (in seconds) its distance \(s(t)\) (in centimeters) from the origin is \(s(t)=\frac{1}{2} a t^{2}+v_{0} t+s_{0}\) for velocity \(v_{0}\) and distance \(s_{0}\) from the origin at \(t=0 .\) If the distances of the particle from the origin at \(t=\frac{1}{2}, t=1,\) and \(t=\frac{3}{2}\) are \(7,11,\) and \(17,\) respectively, find \(a, v_{0+}\) and \(s_{0}\)
Short Answer
Step by step solution
Express Given Conditions as Equations
Convert Equations into a System of Equations
Solve the System of Equations: Eliminate One Variable
Solve the System of Equations: Eliminate Another Variable
Find Acceleration \(a\)
Find Initial Velocity \(v_0\)
Find Initial Position \(s_0\)
Verify the Solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Acceleration
- \( \text{Acceleration} = \frac{\Delta \text{Velocity}}{\Delta \text{Time}} \)
Distance-Time Relationship
- \( s(t) = \frac{1}{2} a t^{2} + v_{0} t + s_{0} \)
System of Equations
- \( \frac{1}{8}a + \frac{1}{2}v_0 + s_0 = 7 \)
- \( \frac{1}{2}a + v_0 + s_0 = 11 \)
- \( \frac{9}{8}a + \frac{3}{2}v_0 + s_0 = 17 \)
Initial Velocity
- \( s(t) = \frac{1}{2} a t^{2} + v_{0} t + s_{0} \)
Initial Position
- \( s_{0} \) is part of \( s(t) = \frac{1}{2} a t^{2} + v_{0} t + s_{0} \)