Question

the distance in feet of a particle from some fixed point is given by the following function, where t is measured in seconds. s(t)=t² + 5t - 5. complete parts (a) through (c). (a) what is the average velocity of the particle over the interval 4 to 5 seconds? (simplify your answer.) (b) what is the average velocity of the particle over the interval 4 to 6 seconds? the average velocity over the interval 4 to 6 is ft/sec. (simplify your answer.) (c) what is the instantaneous velocity of the particle when t = 4? the instantaneous velocity when t = 4 is ft/sec. (simplify your answer.)

Explanation:

Step1: Recall average - velocity formula

The average velocity over the interval $[a,b]$ is given by $\frac{s(b)-s(a)}{b - a}$, where $s(t)=t^{2}+5t - 5$.

Step2: Calculate average velocity for part (a)

For the interval $[4,5]$, $a = 4$ and $b = 5$.
First, find $s(5)$ and $s(4)$:
$s(5)=5^{2}+5\times5 - 5=25 + 25-5=45$.
$s(4)=4^{2}+5\times4 - 5=16 + 20-5=31$.
Then, the average velocity $v_{avg}=\frac{s(5)-s(4)}{5 - 4}=\frac{45 - 31}{1}=14$ ft/sec.

Step3: Calculate average velocity for part (b)

For the interval $[4,6]$, $a = 4$ and $b = 6$.
$s(6)=6^{2}+5\times6 - 5=36+30 - 5=61$.
$s(4)=31$.
The average velocity $v_{avg}=\frac{s(6)-s(4)}{6 - 4}=\frac{61 - 31}{2}=\frac{30}{2}=15$ ft/sec.

Step4: Recall the derivative formula for instantaneous - velocity

The instantaneous velocity $v(t)$ is the derivative of the position function $s(t)$. If $s(t)=t^{2}+5t - 5$, then $v(t)=s^\prime(t)=2t + 5$ using the power rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$.

Step5: Calculate instantaneous velocity for part (c)

When $t = 4$, substitute $t = 4$ into $v(t)$.
$v(4)=2\times4+5=8 + 5=13$ ft/sec.

Answer:

(a) 14 ft/sec
(b) 15 ft/sec
(c) 13 ft/sec