### Physics: Motion of Particles#### Problem Statement:**a.** A particle is undergoing uniform rectilinear motion, and the displacement as a function of time is given by:\[ r(t) = 6t^4 - 2t^3 - 12t^2 + 3t + 3 \]where \( r \) is in meters and \( t \) is in seconds.**b.** What is uniform rectilinear motion?**c.** Find the average acceleration of the particle between \( t = 0 \) and \( t = 2 \) seconds.**d.** When is the acceleration zero?#### Explanation and Calculations:**b. Understanding Uniform Rectilinear Motion:**Uniform rectilinear motion refers to motion along a straight line with constant velocity. This implies that the speed and direction of the particle do not change over time.**c. Calculating the Average Acceleration:**To find the average acceleration, we need to:1. Determine the velocity function \( v(t) \) by differentiating the displacement function \( r(t) \) with respect to time \( t \).2. Determine the acceleration function \( a(t) \) by differentiating the velocity function \( v(t) \) with respect to time \( t \).3. Evaluate the velocity at \( t = 2 \) and \( t = 0 \), and then compute the average acceleration over the interval.\[ r(t) = 6t^4 - 2t^3 - 12t^2 + 3t + 3 \]First derivative to find \( v(t) \) (velocity):\[ v(t) = \frac{d}{dt} r(t) = 24t^3 - 6t^2 - 24t + 3 \]Second derivative to find \( a(t) \) (acceleration):\[ a(t) = \frac{d}{dt} v(t) = 72t^2 - 12t - 24 \]Evaluate \( v(t) \) at \( t = 2 \) and \( t = 0 \):\[ v(2) = 24(2)^3 - 6(2)^2 - 24(2) + 3 = 192 - 24 - 48 + 3 = 123 \, \text{m/s}

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Description: ### Physics: Motion of Particles#### Problem Statement:**a.** A particle is undergoing uniform rectilinear motion, and the displacement as a function of time is given by:\[ r(t) = 6t^4 - 2t^3 - 12t^2 + 3t + 3 \]where \( r \) is in meters and \( t \)

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a. A particle is undergoing a uniform rectilinear motion and the displacement as a function of time is given by r(t) = 6t4- 2t³ -12t²+3t+3 where r is in meters and t is in seconds. b. What is a uniform rectilinear motion? c. Find the average acceleration of the particle between t = 0 and t = 2s d. When is the acceleration zero?

a. A particle is undergoing a uniform rectilinear motion and the displacement as a function of time is given by r(t) = 6t4- 2t³ -12t²+3t+3 where r is in meters and t is in seconds. b. What is a uniform rectilinear motion? c. Find the average acceleration of the particle between t = 0 and t = 2s d. When is the acceleration zero?

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Concept explainers

Displacement, Velocity and Acceleration

In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.

Linear Displacement

The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.

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### Physics: Motion of Particles

#### Problem Statement:
**a.** A particle is undergoing uniform rectilinear motion, and the displacement as a function of time is given by:
\[ r(t) = 6t^4 - 2t^3 - 12t^2 + 3t + 3 \]
where \( r \) is in meters and \( t \) is in seconds.

**b.** What is uniform rectilinear motion?

**c.** Find the average acceleration of the particle between \( t = 0 \) and \( t = 2 \) seconds.

**d.** When is the acceleration zero?

#### Explanation and Calculations:

**b. Understanding Uniform Rectilinear Motion:**
Uniform rectilinear motion refers to motion along a straight line with constant velocity. This implies that the speed and direction of the particle do not change over time.

**c. Calculating the Average Acceleration:**
To find the average acceleration, we need to:
1. Determine the velocity function \( v(t) \) by differentiating the displacement function \( r(t) \) with respect to time \( t \).
2. Determine the acceleration function \( a(t) \) by differentiating the velocity function \( v(t) \) with respect to time \( t \).
3. Evaluate the velocity at \( t = 2 \) and \( t = 0 \), and then compute the average acceleration over the interval.

\[ r(t) = 6t^4 - 2t^3 - 12t^2 + 3t + 3 \]

First derivative to find \( v(t) \) (velocity):
\[ v(t) = \frac{d}{dt} r(t) = 24t^3 - 6t^2 - 24t + 3 \]

Second derivative to find \( a(t) \) (acceleration):
\[ a(t) = \frac{d}{dt} v(t) = 72t^2 - 12t - 24 \]

Evaluate \( v(t) \) at \( t = 2 \) and \( t = 0 \):
\[ v(2) = 24(2)^3 - 6(2)^2 - 24(2) + 3 = 192 - 24 - 48 + 3 = 123 \, \text{m/s}

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