Question

suppose that the position of a particle is given by $s(t)=7t^{3}+5t + 9$.
(a) find the velocity at time $t$.
$v(t)=\frac{m}{s}$
(b) find the velocity at time $t = 3$ seconds.
$\frac{m}{s}$
(c) find the acceleration at time $t$.
$a(t)=\frac{m}{s^{2}}$
(d) find the acceleration at time $t = 3$ seconds.
$\frac{m}{s^{2}}$
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Explanation:

Step1: Find the velocity function

The velocity \(v(t)\) is the derivative of the position function \(s(t)\).
Using the power rule \(\frac{d}{dt}(t^n)=nt^{n - 1}\), for \(s(t)=7t^{3}+5t + 9\), we have:
\(v(t)=\frac{d}{dt}(7t^{3}+5t + 9)=7\times3t^{2}+5\times1t^{0}+0\)
\(v(t)=21t^{2}+5\)

Step2: Find the velocity at \(t = 3\)

Substitute \(t = 3\) into \(v(t)\):
\(v(3)=21\times(3)^{2}+5=21\times9 + 5=189+5=194\)

Step3: Find the acceleration function

The acceleration \(a(t)\) is the derivative of the velocity function \(v(t)\).
Since \(v(t)=21t^{2}+5\), using the power rule \(\frac{d}{dt}(t^n)=nt^{n - 1}\), we get:
\(a(t)=\frac{d}{dt}(21t^{2}+5)=21\times2t^{1}+0\)
\(a(t)=42t\)

Step4: Find the acceleration at \(t = 3\)

Substitute \(t = 3\) into \(a(t)\):
\(a(3)=42\times3=126\)

Answer:

a. \(v(t)=21t^{2}+5\)
b. \(194\)
c. \(a(t)=42t\)
d. \(126\)