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 \(x_1(t)\)): $$a_1(t) = \frac{dv_1}{dt} = \frac{d}{dt} (-8.1 t^2) = -8.1 \times 2t^{2-1} = -16.2 t$$
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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
Consider the motion of a point on a circular trajectory. The acceleration in a linear motion (a) and the acceleration in angular motion (α), are related as : (Take r as the radius of circular trajectory)
A body of mass 10 kg moving with a velocity of 1 m/s is acted upon by a force of 50 N for two seconds. The final velocity will be:
A ball is dropped on a smooth horizontal surface from height ‘h’. What will be the height of rebounce after second impact, if coefficient of restitution between ball and surface is ‘e’?
If water in a stream is flowing with a velocity of 20 kmph and a boat is travelling from one bank to another bank, if the velocity of boat in a direction perpendicular to direction of stream is 20 kmph and width of the stream is 2km, then the time taken and the angle at which boat makes with the direction stream is,
If a train of length 1000 m is cruising at a speed of 90 kmph and crosses a bridge of length 1000 m, what time does it take to completely pass the bridge?
