Question

1. suppose a body moves on a coordinate line such that it has a position in meters given by s(t)=t² - 6t + 3 at t seconds on the interval 0 ≤ t ≤ 5

a) find the displacement of the body on the given time interval.
b) find the average velocity of the body on the given time interval.
c) find the speed of the body at the endpoints of the interval.
d) find the acceleration of the body at the endpoints of the interval.
e) when, if ever, does the body change direction during the time interval?

Explanation:

Step1: Find displacement

Displacement is $s(5)-s(0)$. First, find $s(5)$ and $s(0)$.
\[s(5)=5^{2}-6\times5 + 3=25-30 + 3=-2\]
\[s(0)=0^{2}-6\times0 + 3=3\]
Displacement $=s(5)-s(0)=-2 - 3=-5$ meters.

Step2: Find average velocity

Average velocity $v_{avg}=\frac{s(5)-s(0)}{5 - 0}$. We know $s(5)-s(0)=-5$, so $v_{avg}=\frac{-5}{5}=-1$ m/s.

Step3: Find velocity function

Velocity $v(t)=s^\prime(t)$. Differentiate $s(t)=t^{2}-6t + 3$ using power - rule. $v(t)=2t-6$.

Step4: Find speed at endpoints

At $t = 0$, $v(0)=2\times0-6=-6$, speed is $|v(0)| = 6$ m/s. At $t = 5$, $v(5)=2\times5-6 = 4$, speed is $|v(5)| = 4$ m/s.

Step5: Find acceleration function

Acceleration $a(t)=v^\prime(t)$. Differentiate $v(t)=2t - 6$. $a(t)=2$ m/s². At $t = 0$ and $t = 5$, $a(0)=a(5)=2$ m/s².

Step6: Find when body changes direction

The body changes direction when $v(t)=0$. Set $v(t)=2t-6 = 0$. Solve for $t$: $2t=6$, $t = 3$ seconds.

Answer:

a) Displacement: -5 meters
b) Average velocity: -1 m/s
c) Speed at $t = 0$: 6 m/s, Speed at $t = 5$: 4 m/s
d) Acceleration at $t = 0$ and $t = 5$: 2 m/s²
e) The body changes direction at $t = 3$ seconds