suppose that the position of a particle is given by $s(t)=7t^{3}+5t + 9$.\n(a) find the velocity at time…

suppose that the position of a particle is given by $s(t)=7t^{3}+5t + 9$.\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 $s(t)=7t^{3}+5t + 9$.\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(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)