Josh Sizemore

2021-12-26

Suppose that the position of a particle is given by $s\left(t\right)=4{t}^{3}+5t+9$.
(a) Find the velocity at time t.
$v\left(t\right)=?\frac{m}{c}$
(b) Find the velocity at time $t=-3\mathrm{sec}onds$.
_____ $\frac{m}{s}$
(c) Find the acceleration at time t.
$a\left(t\right)=?\frac{m}{{s}^{2}}$
(d) Find the acceleration at time $t=3\mathrm{sec}onds$.
_____ $\frac{m}{{s}^{2}}$

Thomas Lynn

Step 1 Given
Position of a particle is given by $s\left(t\right)=4{t}^{3}+5t+9$
Step 2 (a) Velocity at time t
Given: Position $s\left(t\right)=4{t}^{3}+5t+9$
Velocity $v\left(t\right)$ is given by: $v\left(t\right)=\frac{d\left[s\left(t\right)\right]}{dt}$
$v\left(t\right)=\frac{d\left[s\left(t\right)\right]}{dt}$
$v\left(t\right)=\frac{d\left(4{t}^{3}+5t+9\right)}{dt}$
$4\left(3\right){t}^{3}-1+5\left(1\right)+0$
$=12{t}^{2}+5$
Therefore, velocity at time t is: $v\left(t\right)=\left(12{t}^{2}+5\right)\frac{m}{s}$

Joseph Lewis

Step 3 (b) Velocity at time $t=3$ seconds
Velocity at time t is: $v\left(t\right)=12{t}^{2}+5$
Therefore, velocity at time $t=3$ seconds is given by
$v\left(t\right)=12{t}^{2}+5$
$v\left(3\right)=12{\left(3\right)}^{2}+5$
$=12\left(9\right)+5$
$=108+5$
$=113$
Therefore, velocity at time $t=3$ seconds is:
$v=113\frac{m}{s}$

karton

Step 4 (c) Acceleration at time t
Velocity at time t is: $v\left(t\right)=12{t}^{2}+5$
Acceleration a(t) is given by: $a\left(t\right)=\frac{d\left[v\left(t\right)\right]}{dt}$
$a\left(t\right)=\frac{d\left[v\left(t\right)\right]}{dt}$
$a\left(t\right)=\frac{d\left(12{t}^{2}+5\right)}{dt}$
$=12\left(2\right){t}^{2-1}+0$
=24t
Therefore, acceleration at time t is: $a\left(t\right)=24t\frac{m}{{s}^{2}}$
Step 5 (d) Acceleration at time t=3 seconds
Acceleration at time t is: a(t)=24t
Therefore, acceleration at time t=3 seconds is given by
a(t)=24t
a(3)=24(3)
=72
Therefore, acceleration at time t=3 seconds is: $a=72\frac{m}{{s}^{2}}$

xleb123

(a) To find the velocity at time t, we need to differentiate the position function s(t) with respect to time. Using the power rule for differentiation, we have:
$v\left(t\right)=\frac{ds\left(t\right)}{dt}=\frac{d}{dt}\left(4{t}^{3}+5t+9\right)=12{t}^{2}+5$ $\frac{m}{s}$
(b) To find the velocity at time t = -3 seconds, we substitute t = -3 into the velocity function:
$v\left(-3\right)=12\left(-3{\right)}^{2}+5=108+5=113$ $\frac{m}{s}$
(c) To find the acceleration at time t, we need to differentiate the velocity function v(t) with respect to time:
$a\left(t\right)=\frac{dv\left(t\right)}{dt}=\frac{d}{dt}\left(12{t}^{2}+5\right)=24t$ $\frac{m}{{s}^{2}}$
(d) To find the acceleration at time t = 3 seconds, we substitute t = 3 into the acceleration function:
$a\left(3\right)=24\left(3\right)=72$ $\frac{m}{{s}^{2}}$

Andre BalkonE

Step 1:
(a) To find the velocity at time t, we need to differentiate the position function s(t) with respect to time. Using the power rule of differentiation, we obtain:
$v\left(t\right)=\frac{ds}{dt}=\frac{d}{dt}\left(4{t}^{3}+5t+9\right)$
Taking the derivative of each term separately, we have:
$v\left(t\right)=\frac{d}{dt}\left(4{t}^{3}\right)+\frac{d}{dt}\left(5t\right)+\frac{d}{dt}\left(9\right)$
Applying the power rule, we get:
$v\left(t\right)=12{t}^{2}+5$
Therefore, the velocity at time t is given by $v\left(t\right)=12{t}^{2}+5\frac{m}{s}$.
Step 2:
(b) To find the velocity at time t = -3 seconds, we substitute t = -3 into the velocity function we derived in part (a). Plugging in t = -3, we have:
$v\left(-3\right)=12\left(-3{\right)}^{2}+5$
Simplifying the expression:
$v\left(-3\right)=12\left(9\right)+5$
$v\left(-3\right)=108+5$
$v\left(-3\right)=113\frac{m}{s}$
Therefore, the velocity at time t = -3 seconds is $v\left(-3\right)=113\frac{m}{s}$.
Step 3:
(c) To find the acceleration at time t, we need to differentiate the velocity function $v\left(t\right)$ with respect to time. Taking the derivative of $v\left(t\right)$, we get:
$a\left(t\right)=\frac{dv}{dt}=\frac{d}{dt}\left(12{t}^{2}+5\right)$
Differentiating each term separately, we obtain:
$a\left(t\right)=\frac{d}{dt}\left(12{t}^{2}\right)+\frac{d}{dt}\left(5\right)$
Using the power rule, we have:
$a\left(t\right)=24t$
Thus, the acceleration at time t is given by $a\left(t\right)=24t\frac{m}{{s}^{2}}$.
Step 4:
(d) To find the acceleration at time t = 3 seconds, we substitute t = 3 into the acceleration function we derived in part (c). Substituting t = 3, we have:
$a\left(3\right)=24\left(3\right)$
$a\left(3\right)=72\frac{m}{{s}^{2}}$
Therefore, the acceleration at time t = 3 seconds is $a\left(3\right)=72\frac{m}{{s}^{2}}$.

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