A particle moves in a straight line. The particle is initially at rest at the origin. The position-time function of the particle is x(t) = 2.7t^2 - 0.27t^3 for 0 ≤ t ≤ 10. Determin... A particle moves in a straight line. The particle is initially at rest at the origin. The position-time function of the particle is x(t) = 2.7t^2 - 0.27t^3 for 0 ≤ t ≤ 10. Determine (a) the average velocity of the particle between t = 2.0 s and t = 4.0 s, (b) the velocity-time function, (c) the average acceleration of the particle between t = 4.0 s and t = 6.0 s, (d) the maximum velocity of the particle in this time period.
Understand the Problem
The question is asking for calculations related to the motion of a particle defined by a position-time function. Specifically, it requires determining the average velocity between given time intervals, the velocity-time function, the average acceleration over another interval, and the maximum velocity within that time period.
Answer
(a) $8.64 \, \text{m/s}$ (b) $v(t) = 5.4t - 0.81t^2$ (c) $-2.7 \, \text{m/s}^2$ (d) $8.64 \, \text{m/s}$
Answer for screen readers
(a) (8.64 , \text{m/s} )
(b) ( v(t) = 5.4t - 0.81t^2 )
(c) (-2.7 , \text{m/s}^2 )
(d) (8.64 , \text{m/s} )
Steps to Solve
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Calculate the Position at Given Times To find the average velocity between (t = 2.0) s and (t = 4.0) s, we first calculate the position of the particle at these times using the position function (x(t) = 2.7t^2 - 0.27t^3).
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For (t = 2.0): $$ x(2.0) = 2.7(2.0)^2 - 0.27(2.0)^3 = 2.7(4) - 0.27(8) = 10.8 - 2.16 = 8.64 \text{ m} $$
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For (t = 4.0): $$ x(4.0) = 2.7(4.0)^2 - 0.27(4.0)^3 = 2.7(16) - 0.27(64) = 43.2 - 17.28 = 25.92 \text{ m} $$
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Calculate the Average Velocity Average velocity (v_{\text{avg}}) over the interval ([t_1, t_2]) can be calculated using the formula: $$ v_{\text{avg}} = \frac{x(t_2) - x(t_1)}{t_2 - t_1} $$ Substituting the values we calculated: $$ v_{\text{avg}} = \frac{25.92 - 8.64}{4.0 - 2.0} = \frac{17.28}{2.0} = 8.64 \text{ m/s} $$
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Determine the Velocity-Time Function The velocity (v(t)) is found by differentiating the position function (x(t)): $$ v(t) = \frac{dx}{dt} = \frac{d}{dt}(2.7t^2 - 0.27t^3) $$ Using the power rule: $$ v(t) = 5.4t - 0.81t^2 $$
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Calculate the Average Acceleration Average acceleration (a_{\text{avg}}) over the interval ([4.0, 6.0]) can be calculated similarly: First, find (v(4)) and (v(6)):
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For (t = 4.0): $$ v(4.0) = 5.4(4.0) - 0.81(4.0)^2 = 21.6 - 12.96 = 8.64 \text{ m/s} $$
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For (t = 6.0): $$ v(6.0) = 5.4(6.0) - 0.81(6.0)^2 = 32.4 - 29.16 = 3.24 \text{ m/s} $$
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Now, average acceleration is calculated as: $$ a_{\text{avg}} = \frac{v(6.0) - v(4.0)}{6.0 - 4.0} = \frac{3.24 - 8.64}{2.0} = \frac{-5.4}{2.0} = -2.7 \text{ m/s}^2 $$
- Find the Maximum Velocity
To find the maximum velocity, check the critical points by setting (v(t) = 0):
$$
5.4t - 0.81t^2 = 0
$$
Factoring:
$$
t(5.4 - 0.81t) = 0
$$
Critical points are (t = 0) and (t = \frac{5.4}{0.81} \approx 6.67) (outside the interval). Evaluate at (t = 4) and (t = 6).
From earlier results:
- (v(4.0) = 8.64 \text{ m/s})
- (v(6.0) = 3.24 \text{ m/s})
The maximum velocity in the interval is (8.64 \text{ m/s}).
(a) (8.64 , \text{m/s} )
(b) ( v(t) = 5.4t - 0.81t^2 )
(c) (-2.7 , \text{m/s}^2 )
(d) (8.64 , \text{m/s} )
More Information
The calculations involved finding position values at specified times to derive average velocity, differentiating the position-time function to obtain velocity, and then applying the average acceleration formula. The maximum velocity was determined by evaluating the function within the time limits.
Tips
- Forgetting to differentiate correctly when finding the velocity function.
- Not accurately computing the values during position evaluation, leading to wrong average velocity calculations.
- Ignoring the proper interval when determining maximum velocity.
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