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Answer :
We are given the position function
[tex]$$
x(t) = 2.75t^2 - 2.00t + 3.00,
$$[/tex]
with [tex]$x$[/tex] in meters and [tex]$t$[/tex] in seconds.
Below are the steps to find the required quantities.
────────────────────────
(a) Average Speed between [tex]$t = 1.70\text{ s}$[/tex] and [tex]$t = 3.40\text{ s}$[/tex]
1. First, compute the positions at the two times:
[tex]\[
x(1.70) = 2.75(1.70)^2 - 2.00(1.70) + 3.00 = 7.5475\text{ m},
\][/tex]
[tex]\[
x(3.40) = 2.75(3.40)^2 - 2.00(3.40) + 3.00 = 27.99\text{ m}.
\][/tex]
2. The displacement is
[tex]\[
\Delta x = x(3.40) - x(1.70) = 27.99 - 7.5475 = 20.4425\text{ m}.
\][/tex]
3. The time interval is
[tex]\[
\Delta t = 3.40 - 1.70 = 1.70\text{ s}.
\][/tex]
4. Therefore, the average speed is
[tex]\[
v_{\text{avg}} = \frac{\Delta x}{\Delta t} \approx \frac{20.4425}{1.70} \approx 12.025\text{ m/s}.
\][/tex]
────────────────────────
(b) Instantaneous Speed at Specific Times
1. The instantaneous velocity is the derivative of [tex]$x(t)$[/tex]. Differentiating, we get
[tex]\[
v(t) = \frac{dx}{dt} = 2(2.75)t - 2.00 = 5.5t - 2.00.
\][/tex]
2. At [tex]$t = 1.70\text{ s}$[/tex]:
[tex]\[
v(1.70) = 5.5(1.70) - 2.00 = 9.35 - 2.00 = 7.35\text{ m/s}.
\][/tex]
3. At [tex]$t = 3.40\text{ s}$[/tex]:
[tex]\[
v(3.40) = 5.5(3.40) - 2.00 = 18.70 - 2.00 = 16.70\text{ m/s}.
\][/tex]
────────────────────────
(c) Average Acceleration between [tex]$t = 1.70\text{ s}$[/tex] and [tex]$t = 3.40\text{ s}$[/tex]
1. The average acceleration is given by
[tex]\[
a_{\text{avg}} = \frac{v(3.40) - v(1.70)}{3.40 - 1.70}.
\][/tex]
2. Substituting the computed velocities:
[tex]\[
a_{\text{avg}} = \frac{16.70 - 7.35}{1.70} = \frac{9.35}{1.70} \approx 5.5\text{ m/s}^2.
\][/tex]
────────────────────────
(d) Instantaneous Acceleration at Specific Times
1. The instantaneous acceleration is the derivative of the velocity function. Since
[tex]\[
v(t) = 5.5t - 2.00,
\][/tex]
its derivative is
[tex]\[
a(t) = \frac{dv}{dt} = 5.5.
\][/tex]
2. Hence, the instantaneous acceleration at any time is
[tex]\[
a(1.70) = a(3.40) = 5.5\text{ m/s}^2.
\][/tex]
────────────────────────
(e) Time When the Object is at Rest
1. The object is at rest when [tex]$v(t) = 0$[/tex]. Set up the equation:
[tex]\[
5.5t - 2.00 = 0.
\][/tex]
2. Solve for [tex]$t$[/tex]:
[tex]\[
5.5t = 2.00 \quad \Longrightarrow \quad t = \frac{2.00}{5.5} \approx 0.3636\text{ s}.
\][/tex]
────────────────────────
Summary of Results:
- (a) Average speed between [tex]$t=1.70\text{ s}$[/tex] and [tex]$t=3.40\text{ s}$[/tex]: [tex]$\boxed{12.025\ \text{m/s}}$[/tex].
- (b) Instantaneous speed at [tex]$t=1.70\text{ s}$[/tex]: [tex]$\boxed{7.35\ \text{m/s}}$[/tex]; at [tex]$t=3.40\text{ s}$[/tex]: [tex]$\boxed{16.70\ \text{m/s}}$[/tex].
- (c) Average acceleration between [tex]$t=1.70\text{ s}$[/tex] and [tex]$t=3.40\text{ s}$[/tex]: [tex]$\boxed{5.5\ \text{m/s}^2}$[/tex].
- (d) Instantaneous acceleration at both [tex]$t=1.70\text{ s}$[/tex] and [tex]$t=3.40\text{ s}$[/tex]: [tex]$\boxed{5.5\ \text{m/s}^2}$[/tex].
- (e) The object is at rest at [tex]$t \approx \boxed{0.3636\ \text{s}}$[/tex].
[tex]$$
x(t) = 2.75t^2 - 2.00t + 3.00,
$$[/tex]
with [tex]$x$[/tex] in meters and [tex]$t$[/tex] in seconds.
Below are the steps to find the required quantities.
────────────────────────
(a) Average Speed between [tex]$t = 1.70\text{ s}$[/tex] and [tex]$t = 3.40\text{ s}$[/tex]
1. First, compute the positions at the two times:
[tex]\[
x(1.70) = 2.75(1.70)^2 - 2.00(1.70) + 3.00 = 7.5475\text{ m},
\][/tex]
[tex]\[
x(3.40) = 2.75(3.40)^2 - 2.00(3.40) + 3.00 = 27.99\text{ m}.
\][/tex]
2. The displacement is
[tex]\[
\Delta x = x(3.40) - x(1.70) = 27.99 - 7.5475 = 20.4425\text{ m}.
\][/tex]
3. The time interval is
[tex]\[
\Delta t = 3.40 - 1.70 = 1.70\text{ s}.
\][/tex]
4. Therefore, the average speed is
[tex]\[
v_{\text{avg}} = \frac{\Delta x}{\Delta t} \approx \frac{20.4425}{1.70} \approx 12.025\text{ m/s}.
\][/tex]
────────────────────────
(b) Instantaneous Speed at Specific Times
1. The instantaneous velocity is the derivative of [tex]$x(t)$[/tex]. Differentiating, we get
[tex]\[
v(t) = \frac{dx}{dt} = 2(2.75)t - 2.00 = 5.5t - 2.00.
\][/tex]
2. At [tex]$t = 1.70\text{ s}$[/tex]:
[tex]\[
v(1.70) = 5.5(1.70) - 2.00 = 9.35 - 2.00 = 7.35\text{ m/s}.
\][/tex]
3. At [tex]$t = 3.40\text{ s}$[/tex]:
[tex]\[
v(3.40) = 5.5(3.40) - 2.00 = 18.70 - 2.00 = 16.70\text{ m/s}.
\][/tex]
────────────────────────
(c) Average Acceleration between [tex]$t = 1.70\text{ s}$[/tex] and [tex]$t = 3.40\text{ s}$[/tex]
1. The average acceleration is given by
[tex]\[
a_{\text{avg}} = \frac{v(3.40) - v(1.70)}{3.40 - 1.70}.
\][/tex]
2. Substituting the computed velocities:
[tex]\[
a_{\text{avg}} = \frac{16.70 - 7.35}{1.70} = \frac{9.35}{1.70} \approx 5.5\text{ m/s}^2.
\][/tex]
────────────────────────
(d) Instantaneous Acceleration at Specific Times
1. The instantaneous acceleration is the derivative of the velocity function. Since
[tex]\[
v(t) = 5.5t - 2.00,
\][/tex]
its derivative is
[tex]\[
a(t) = \frac{dv}{dt} = 5.5.
\][/tex]
2. Hence, the instantaneous acceleration at any time is
[tex]\[
a(1.70) = a(3.40) = 5.5\text{ m/s}^2.
\][/tex]
────────────────────────
(e) Time When the Object is at Rest
1. The object is at rest when [tex]$v(t) = 0$[/tex]. Set up the equation:
[tex]\[
5.5t - 2.00 = 0.
\][/tex]
2. Solve for [tex]$t$[/tex]:
[tex]\[
5.5t = 2.00 \quad \Longrightarrow \quad t = \frac{2.00}{5.5} \approx 0.3636\text{ s}.
\][/tex]
────────────────────────
Summary of Results:
- (a) Average speed between [tex]$t=1.70\text{ s}$[/tex] and [tex]$t=3.40\text{ s}$[/tex]: [tex]$\boxed{12.025\ \text{m/s}}$[/tex].
- (b) Instantaneous speed at [tex]$t=1.70\text{ s}$[/tex]: [tex]$\boxed{7.35\ \text{m/s}}$[/tex]; at [tex]$t=3.40\text{ s}$[/tex]: [tex]$\boxed{16.70\ \text{m/s}}$[/tex].
- (c) Average acceleration between [tex]$t=1.70\text{ s}$[/tex] and [tex]$t=3.40\text{ s}$[/tex]: [tex]$\boxed{5.5\ \text{m/s}^2}$[/tex].
- (d) Instantaneous acceleration at both [tex]$t=1.70\text{ s}$[/tex] and [tex]$t=3.40\text{ s}$[/tex]: [tex]$\boxed{5.5\ \text{m/s}^2}$[/tex].
- (e) The object is at rest at [tex]$t \approx \boxed{0.3636\ \text{s}}$[/tex].
Thank you for reading the article An object moves along the tex x tex axis according to the equation tex x 2 75 t 2 2 00 t 3 00 tex. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
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Rewritten by : Jeany
