Question

suppose that the position of a particle is given by ( s = f(t) = 7t^{3}+6t + 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}} )
question help: video message instructor
submit question jump to answer

Explanation:

Step1: Find the velocity function

The velocity \(v(t)\) is the derivative of the position function \(s = f(t)=7t^{3}+6t + 9\).
Using the power rule \(\frac{d}{dt}(t^{n})=nt^{n - 1}\), we have:
\(v(t)=\frac{ds}{dt}=\frac{d}{dt}(7t^{3}+6t + 9)\)
\(v(t)=7\times3t^{2}+6\times1t^{0}+0\)
\(v(t)=21t^{2}+6\)

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

Substitute \(t = 3\) into \(v(t)\):
\(v(3)=21\times(3)^{2}+6\)
\(v(3)=21\times9 + 6\)
\(v(3)=189+6\)
\(v(3)=195\)

Step3: Find the acceleration function

The acceleration \(a(t)\) is the derivative of the velocity function \(v(t)=21t^{2}+6\).
Using the power rule \(\frac{d}{dt}(t^{n})=nt^{n - 1}\), we have:
\(a(t)=\frac{dv}{dt}=\frac{d}{dt}(21t^{2}+6)\)
\(a(t)=21\times2t^{1}+0\)
\(a(t)=42t\)

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

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

Answer:

(a) \(v(t)=21t^{2}+6\) \(\frac{m}{s}\)
(b) \(195\) \(\frac{m}{s}\)
(c) \(a(t)=42t\) \(\frac{m}{s^{2}}\)
(d) \(126\) \(\frac{m}{s^{2}}\)