1
Question
A particle executes the motion described by
\(x(t)=x_0(1-e^{\gamma t})\); \(t\geq0,x_0>0\).
Where does the particle start and with what velocity?
\(x(t)=x_0(1-e^{\gamma t})\); \(t\geq0,x_0>0\).
Where does the particle start and with what velocity?
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Solution
Step 1:Find velocity of the particle.
Formula used: \(𝑣=\dfrac{𝑑𝑥}{𝑑𝑡}\)
Given,
Displacement as a function of time,
\(x(t)=x_0(1-e^{-\gamma t})\)
By differentiating this equation with respect to the time
\(v(t)=\dfrac{dx}{dt}\)
\(v(t)=x_0\gamma{e}^{-\gamma t}\)
Step 2:Find position of the particle when it starts motion.
Particle starts its motion at time \(𝑡=0\),
\(x(t)=x_0(1-e^{-0})\)
\(x=x_0(1-1)\)
\(x=0\)
Step 3: Find velocity of the particle when it starts motion. Velocity of the particle at time \(𝑡=0\),
\(v(t)=x_0\gamma{e}^{-\gamma(0)}\)
\(v(t)=x_0\gamma(1)\)
\(=\gamma x_0\)
Final answer: \(x=0;v=\gamma x_0\)
Formula used: \(𝑣=\dfrac{𝑑𝑥}{𝑑𝑡}\)
Given,
Displacement as a function of time,
\(x(t)=x_0(1-e^{-\gamma t})\)
By differentiating this equation with respect to the time
\(v(t)=\dfrac{dx}{dt}\)
\(v(t)=x_0\gamma{e}^{-\gamma t}\)
Step 2:Find position of the particle when it starts motion.
Particle starts its motion at time \(𝑡=0\),
\(x(t)=x_0(1-e^{-0})\)
\(x=x_0(1-1)\)
\(x=0\)
Step 3: Find velocity of the particle when it starts motion. Velocity of the particle at time \(𝑡=0\),
\(v(t)=x_0\gamma{e}^{-\gamma(0)}\)
\(v(t)=x_0\gamma(1)\)
\(=\gamma x_0\)
Final answer: \(x=0;v=\gamma x_0\)
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