suppose that the position of a particle is given by ( s = f(t) = 7t^{3}+6t + 9 ).\n(a) find the velocity at…

suppose that the position of a particle is given by ( s = f(t) = 7t^{3}+6t + 9 ).\n(a) find the velocity at time ( t ).\n( v(t)=\frac{m}{s} )\n(b) find the velocity at time ( t = 3 ) seconds.\n( \frac{m}{s} )\n(c) find the acceleration at time ( t ).\n( a(t)=\frac{m}{s^{2}} )\n(d) find the acceleration at time ( t = 3 ) seconds.\n( \frac{m}{s^{2}} )\nquestion help: video message instructor\nsubmit question jump to answer

suppose that the position of a particle is given by ( s = f(t) = 7t^{3}+6t + 9 ).\n(a) find the velocity at time ( t ).\n( v(t)=\frac{m}{s} )\n(b) find the velocity at time ( t = 3 ) seconds.\n( \frac{m}{s} )\n(c) find the acceleration at time ( t ).\n( a(t)=\frac{m}{s^{2}} )\n(d) find the acceleration at time ( t = 3 ) seconds.\n( \frac{m}{s^{2}} )\nquestion help: video message instructor\nsubmit question jump to answer

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}})