suppose that the position of a particle is given by ( s(t)=2 t^{3}+7 t + 9 ).\n(a) find the velocity at time…

suppose that the position of a particle is given by ( s(t)=2 t^{3}+7 t + 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(t)=2 t^{3}+7 t + 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: Recall the velocity formula

Velocity (v(t)) is the derivative of position function (s(t)). Using the power rule (\frac{d}{dt}(x^n)=nx^{n - 1}), for (s(t)=2t^{3}+7t + 9), (v(t)=s^\prime(t)). (v(t)=\frac{d}{dt}(2t^{3})+\frac{d}{dt}(7t)+\frac{d}{dt}(9)) (v(t)=2\times3t^{2}+7\times1t^{0}+0) (v(t)=6t^{2}+7)

Step2: Find (v(3))

Substitute (t = 3) into (v(t)). (v(3)=6\times(3)^{2}+7) (v(3)=6\times9 + 7) (v(3)=54+7=61)

Step3: Recall the acceleration formula

Acceleration (a(t)) is the derivative of velocity function (v(t)). Since (v(t)=6t^{2}+7), (a(t)=v^\prime(t)). Using the power rule (\frac{d}{dt}(x^n)=nx^{n - 1}), (a(t)=\frac{d}{dt}(6t^{2})+\frac{d}{dt}(7)) (a(t)=6\times2t^{1}+0) (a(t)=12t)

Step4: Find (a(3))

Substitute (t = 3) into (a(t)). (a(3)=12\times3=36)

Answer:

(a) (v(t)=6t^{2}+7) (\frac{m}{s}) (b) (61) (\frac{m}{s}) (c) (a(t)=12t) (\frac{m}{s^{2}}) (d) (36) (\frac{m}{s^{2}})