suppose the position of a particle, s(t) is given by s(t)=t^2 - 10t + 28 feet, where t is time in seconds…

suppose the position of a particle, s(t) is given by s(t)=t^2 - 10t + 28 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 - 10t + 28 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: Find the derivative for velocity function

The derivative of $s(t)=t^{2}-10t + 28$ using the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$. $v(t)=s'(t)=\frac{d}{dt}(t^{2}-10t + 28)=2t-10$

Step2: Evaluate velocity at $t = 3$

Substitute $t = 3$ into $v(t)$. $v(3)=s'(3)=2\times3-10=6 - 10=-4$ feet per second

Step3: Find when the particle is at rest

Set $v(t)=s'(t)=0$. $2t-10 = 0$ Add 10 to both sides: $2t=10$ Divide both sides by 2: $t = 5$ seconds

Answer:

(a) $2t-10$ (b) -4 feet per second (c) 5 seconds