Given the equation of motion for an object moving along the x-axis: $x(t) = 4.00t^2 - 2.00t + 3.00$ where $x$ is in meters and $t$ is in seconds, we'll determine several key parameters:

(a) Average Speed between $t = 1.50 \, \text{s}$ and $t = 3.40 \, \text{s}$

The average speed is calculated as: $\text{Average speed} = \frac{|x(t_2) - x(t_1)|}{t_2 - t_1}$ where $t_1 = 1.50 \, \text{s}$ and $t_2 = 3.40 \, \text{s}$.

First, find the positions at $t = 1.50 \, \text{s}$ and $t = 3.40 \, \text{s}$: $x(1.50) = 4.00(1.50)^2 - 2.00(1.50) + 3.00$ $x(3.40) = 4.00(3.40)^2 - 2.00(3.40) + 3.00$ Now, compute the average speed.

(b) Instantaneous Speed at $t = 1.50 \, \text{s}$ and $t = 3.40 \, \text{s}$

To find the instantaneous speed, we first need to calculate the velocity by taking the derivative of the position function: $v(t) = \frac{dx}{dt} = \frac{d}{dt}\left(4.00t^2 - 2.00t + 3.00\right)$ $v(t) = 8.00t - 2.00$ At $t = 1.50 \, \text{s}$: $v(1.50) = 8.00(1.50) - 2.00$ At $t = 3.40 \, \text{s}$: $v(3.40) = 8.00(3.40) - 2.00$ Since speed is the magnitude of velocity, the instantaneous speeds will be the absolute values of the velocities at these times.

(c) Average Acceleration between $t = 1.50 \, \text{s}$ and $t = 3.40 \, \text{s}$

The average acceleration is calculated as: $\text{Average acceleration} = \frac{v(t_2) - v(t_1)}{t_2 - t_1}$ Substitute the values of $v(3.40)$ and $v(1.50)$ that were calculated earlier.

(d) Instantaneous Acceleration at $t = 1.50 \, \text{s}$ and $t = 3.40 \, \text{s}$

To find the instantaneous acceleration, we take the derivative of the velocity function: $a(t) = \frac{dv}{dt} = \frac{d}{dt}(8.00t - 2.00)$ $a(t) = 8.00 \, \text{m/s}^2$ Thus, the instantaneous acceleration is constant and equal to $8.00 \, \text{m/s}^2$ at both $t = 1.50 \, \text{s}$ and $t = 3.40 \, \text{s}$.

(e) Time when the object is at rest

The object is at rest when the velocity is zero. Set the velocity equation $v(t) = 8.00t - 2.00$ to zero and solve for $t$: $8.00t - 2.00 = 0$ $t = \frac{2.00}{8.00} = 0.25 \, \text{s}$

Now, let's calculate these values step by step.Here are the results for the problem:

(a) Average speed between $t = 1.50 \, \text{s}$ and $t = 3.40 \, \text{s}$:

$\text{Average speed} = 17.60 \, \text{m/s}$

(b) Instantaneous speed at:

• $t = 1.50 \, \text{s}$: $10.00 \, \text{m/s}$
• $t = 3.40 \, \text{s}$: $25.20 \, \text{m/s}$

(c) Average acceleration between $t = 1.50 \, \text{s}$ and $t = 3.40 \, \text{s}$:

$\text{Average acceleration} = 8.00 \, \text{m/s}^2$

(d) Instantaneous acceleration (constant) at:

• $t = 1.50 \, \text{s}$: $8.00 \, \text{m/s}^2$
• $t = 3.40 \, \text{s}$: $8.00 \, \text{m/s}^2$

(e) The object is at rest at $t = 0.25 \, \text{s}$.

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Would you like further details or clarification on any part of this?

Here are five related questions:

1. How do you calculate average velocity if the motion is non-uniform?
2. What is the difference between speed and velocity?
3. How does acceleration change in uniformly accelerated motion?
4. What physical interpretation can be given to negative acceleration?
5. How do you find displacement from a velocity-time graph?

Tip: Instantaneous speed is always the magnitude of the velocity, while average speed is the total distance divided by total time.