Physics, 06.08.2019 02:30 LikeIke2859

Abody moves on a coordinate line such that it has a position s=f(t) =t^2-−33t++22 on the interval 0≤t≤88, with s in meters and t in seconds. a. find the body's displacement and average velocity for the given time interval. b. find the body's speed and acceleration at the endpoints of the interval. c. when, if ever, during the interval does the body change direction?

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The position equation of moving body on a coordinate line tells the distance of body from initial point in different time interval.

a)The body's displacement and average velocity for the given time interval is 4840 meters and 55 m/s respectively.b)The body's speed at the endpoints of the interval is 33 and 143 seconds, and acceleration at the endpoints of the interval is 2 m/s squaredc) The interval in which the body change direction is at 16.5 seconds.What is position equation?

A position equation of a moving body tells the position or the distance of the body from the initial point in the different time interval.

Given information-

The position equation of the body is,

s=f(t) =t^2-33t+22

The interval for this position is,

$0\leq t \leq 88$

a)The body's displacement and average velocity for the given time interval-

The position of the body when value of time is 0.

$s=f(t) =0^2-33(0)+22\\s=22$

Thus the initial position of the body is 22 meters.

The position of the body when value of time is 88.

$s=f(t) =88^2-33(88)+22\\s=4862 \rm m$

Thus the final position of the body is 4862 meters.

The body's displacement is the difference of final position to the initial position. Thus displacement is,

$d=4862-22\\d=4840\rm m$

The average velocity of the position is the total distance traveled by the body in the total time taken by it. Thus average velocity is,

$v=\dfrac{4840}{88}\\v=55\rm m/s$

Thus the body's displacement and average velocity for the given time interval is 4840 meters and 55 m/s respectively.

b)The body's speed and acceleration at the endpoints of the interval.

The speed of the body is absolute value of the velocity. The speed of the body is the change of position of the body with respect to the time. thus,

$s=|\dfrac{ds}{dt}|\\s=|\dfrac{d(t^2-33t+22)}{dt}|\\s=|{2t-33}|\\$

Endpoint of the interval are 0 and 88.Thus put the values we get,

$\left \{ {{s=33, \;\;\;t=0} \atop {s=143, \;t=88}} \right.$

Acceleration body is the change of the velocity of the body with respect to the time. thus,

$a=\dfrac{dv}{dt}|\\a=\dfrac{d(2t-33)}{dt}\\a={2}\rm m/s^2\\$

Thus the body's speed at the endpoints of the interval is 33 and 143 seconds, and acceleration at the endpoints of the interval is 2 m/s squared

c) The interval in which the body change direction-

Equate the equation of velocity to the zero to find the change of direction of body as,

$2t-33=0\\t=\dfrac{33}{2}\\t=16.5\rm s$

Hence, the interval in which the body change direction is at 16.5 seconds.

Thus,

Learn more about the position equation here;

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Answer from: Coscutt

a.Displacement=4840 meter

Average velocity=55 m/s

b.t=0 then speed=33m/s

when t=88 then speed=143 m/s

Acceleration= $\frac{dv}{dt}=2m/s^2$

c.yes

Explanation:

We are given that a body moves on a coordinate line such that it has a position s=f(t)= t^2-33t+22 Where t is time in sec and s in meters

and t belongs to [0,88]

a.Position when t=0

s=0-0+22=22

when t=88

Then s= (88)^2-33(88)+22 =4862

Displacement=4862-22=4840 meter

Average velocity = $\frac{f(b)-f(a)}{b-a}$

Where a=0 ,b=88,f(88)=4862,f(0)=22

Average velocity= $\frac{4862-22}{88-0}=\frac{4840}{88}$ =55m/s

Average velocity=55 m/s

b.Speed = absolute value of velocity

Speed = $\mid\frac{ds}{dt}\mid=\mid2t-33\mid$

t=0 then speed=33m/s

when t=88 then speed=143 m/s

Acceleration= $\frac{dv}{dt}=2m/s^2$

Hence, acceleration remains constant.

When v=0 then

2t-33=0

t= $\frac{33}{2}=16.5 s$

Acceleration at t=16.5 sec =2 m/s^2

Hence , when v=0 at t=16.5 sec then acceleration does not zero .Therefore, body changes the direction.

Answer from: Quest

Idk just need points

Answer from: Quest

Answerit increases at a steady rate;

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