suppose the position of a particle, s(t) is given by s(t)=t^{2}-5t + 22 feet, where t is time in seconds…

suppose the position of a particle, s(t) is given by s(t)=t^{2}-5t + 22 feet, where t is time in seconds. (a) find the velocity function at time t. s(t)=v(t)= (b) find s(3), the velocity after 3 seconds. include units. (c) find all value(s) of t (with correct units) for which the particle is at rest (i.e., where s(t)=0). if there are no such values, enter none. if there is more than one value, list them separated by commas. t=

suppose the position of a particle, s(t) is given by s(t)=t^{2}-5t + 22 feet, where t is time in seconds. (a) find the velocity function at time t. s(t)=v(t)= (b) find s(3), the velocity after 3 seconds. include units. (c) find all value(s) of t (with correct units) for which the particle is at rest (i.e., where s(t)=0). if there are no such values, enter none. if there is more than one value, list them separated by commas. t=

Answer

Explanation:

Step1: Differentiate to find velocity function

The derivative of (s(t)=t^{2}-5t + 22) using the power - rule ((x^n)^\prime=nx^{n - 1}). [s^\prime(t)=v(t)=\frac{d}{dt}(t^{2}-5t + 22)=2t-5]

Step2: Evaluate (s^\prime(3))

Substitute (t = 3) into (v(t)=2t - 5). [v(3)=2\times3-5=6 - 5=1] The units of velocity are feet per second since (s) is in feet and (t) is in seconds. So (s^\prime(3)=1) feet per second.

Step3: Find when the particle is at rest

Set (s^\prime(t)=0), so (2t-5 = 0). Solve for (t): [2t=5\Rightarrow t=\frac{5}{2}\text{ seconds}]

Answer:

(a) (2t - 5) (b) (1) feet per second (c) (\frac{5}{2}\text{ seconds})