Each of four particles move along an x-axis. Their coordinates (in meters) as functions of time (in seconds) are given by 1) particle 1: x (t) = 3.5 – 2.7 t3 2) particle 2: x (t) = 3.5 + 2.7 t3 3) particle 3: x (t) = 3.5 – 2.7 t2 4) particle 4: x (t) = 3.5 – 3.4t - 2.7 t2 Which of these particles have constant acceleration?
Only (3) and (4)
To determine which of the given particles have constant acceleration, we need to analyze their position functions with respect to time. Acceleration is the second derivative of the position function, \(x(t)\), with respect to time, \(t\). If the resulting acceleration function, \(a(t)\), does not depend on \(t\) (i.e., it's a numerical constant), then the particle has constant acceleration.
Let's examine each particle's motion along the x-axis:
Particle Motion Analysis
-
Particle 1:
The position function is given by \(x_1(t) = 3.5 - 2.7 t^3\).
- Velocity \(v_1(t)\) is the first derivative of \(x_1(t)\) with respect to \(t\): $$v_1(t) = \frac{dx_1}{dt} = \frac{d}{dt} (3.5 - 2.7 t^3) = 0 - 2.7 \times 3t^{3-1} = -8.1 t^2$$
- Acceleration \(a_1(t)\) is the first derivative of \(v_1(t)\) with respect to \(t\) (or the second derivative of \(x_1(t)\)): $$a_1(t) = \frac{dv_1}{dt} = \frac{d}{dt} (-8.1 t^2) = -8.1 \times 2t^{2-1} = -16.2 t$$
-
Particle 2:
The position function is given by \(x_2(t) = 3.5 + 2.7 t^3\).
- Velocity \(v_2(t)\): $$v_2(t) = \frac{dx_2}{dt} = \frac{d}{dt} (3.5 + 2.7 t^3) = 0 + 2.7 \times 3t^{3-1} = 8.1 t^2$$
- Acceleration \(a_2(t)\): $$a_2(t) = \frac{dv_2}{dt} = \frac{d}{dt} (8.1 t^2) = 8.1 \times 2t^{2-1} = 16.2 t$$
-
Particle 3:
The position function is given by \(x_3(t) = 3.5 - 2.7 t^2\).
- Velocity \(v_3(t)\): $$v_3(t) = \frac{dx_3}{dt} = \frac{d}{dt} (3.5 - 2.7 t^2) = 0 - 2.7 \times 2t^{2-1} = -5.4 t$$
- Acceleration \(a_3(t)\): $$a_3(t) = \frac{dv_3}{dt} = \frac{d}{dt} (-5.4 t) = -5.4$$
-
Particle 4:
The position function is given by \(x_4(t) = 3.5 - 3.4t - 2.7 t^2\).
- Velocity \(v_4(t)\): $$v_4(t) = \frac{dx_4}{dt} = \frac{d}{dt} (3.5 - 3.4t - 2.7 t^2) = 0 - 3.4 \times 1t^{1-1} - 2.7 \times 2t^{2-1} = -3.4 - 5.4 t$$
- Acceleration \(a_4(t)\): $$a_4(t) = \frac{dv_4}{dt} = \frac{d}{dt} (-3.4 - 5.4 t) = 0 - 5.4 = -5.4$$
Constant Acceleration Summary
| Particle | Position Function \(x(t)\) | Velocity Function \(v(t)\) | Acceleration Function \(a(t)\) | Constant Acceleration? |
|---|---|---|---|---|
| 1 | \(3.5 - 2.7 t^3\) | \(-8.1 t^2\) | \(-16.2 t\) | No |
| 2 | \(3.5 + 2.7 t^3\) | \(8.1 t^2\) | \(16.2 t\) | No |
| 3 | \(3.5 - 2.7 t^2\) | \(-5.4 t\) | \(-5.4\) | Yes |
| 4 | \(3.5 - 3.4t - 2.7 t^2\) | \(-3.4 - 5.4 t\) | \(-5.4\) | Yes |
Based on our analysis, particles 3 and 4 have acceleration functions that are constant values and do not depend on time \(t\). Therefore, these particles exhibit constant acceleration.
The particles with constant acceleration are only (3) and (4).
Important Questions from Kinematics and Kinetics
What is the coefficient of restitution (e) for elastic impact?
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How much force will be exerted by the floor of the lift on a passenger of 80 kg mass when lift is accelerating downward at 0.81 m/s2?
