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Hitman42 [59]
3 years ago
15

A body moves on a coordinate line such that it has a position sequals​f(t)equalstsquaredminus4tplus3on the interval 0less than o

r equalstless than or equals7​,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?
Physics
1 answer:
Zanzabum3 years ago
3 0

Answer:

A)  Δf = - 49 m, B)  v (7)  = -56 m / s, a = - 8 m / s², C)  t = 0.866 s

Explanation:

A) In this exercise ask to find the displacement and the average velocity, give the function of the movement

           f (t) = - 4t² +3

and the range of motion 0≤ t ≤ 7

the displacement is

for t = 0

           f (0) = 3

for t = 7 s

           f (7) = - 4 7² +3

           f (7) = -46 m

the total displacement is

           Δf = f (7) - f (0)

           Δf = -46 - 3

            Δf = - 49 m

the average speed is defined as the displacement between the time interval

           v = Df / Dt

           v = -49 / 7

           v = - 7 m / s

B) the speed and acceleration of the end points of the motion

         

the speed of defined by

          v = \frac{dx}{dt}

in this case

         v = \frac{df}{dt}

         v = -8t

         

let's calculate

          v (7) = -8 7

          v (7)  = -56 m / s

acceleration is defined by

          a = \frac{dv}{dt}

          a = - 8 m / s²

acceleration is constant throughout the movement

C) the point where the direction changes.

This point is a point where the position goes from positive to negative, the point f = 0

        0 = -4t² +3

        t = √¾

        t = 0.866 s

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Leya [2.2K]

Answer:

A) 0.9844 s

B) x2 = 0.4587 m

C) v = 6.657 m/s

Explanation:

We are given;

Height of take off point above pool; x1 = 1.8 m

Initial take off velocity; u = 3 m/s

Final velocity at highest point before free fall; v = 0 m/s

B) To find the highest point above the board her feet reaches means the distance from take off to the top of the motion just before free fall.

Thus, we will be using equation of motion and we have;

v² = u² + 2gs

Now, let s = x2 which will be the distance between take off and the top before free fall.

So;

v² = u² + 2g(x2)

Now,since the motion is against gravity, g will be negative.

Thus;

v² = u² + 2(-9.81)(x2)

Plugging in the relevant values to give;

0² = 3² - (19.62x2)

19.62(x2) = 9

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A) We want to find how long her feet is in air.. It means we want to find out the time to get to a distance of x1 and also the time to achieve the distance (x1 + x2) on free-fall.

Thus, using equation of motion;

v = u + gt

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0 = 3 - 9.81t1

9.81t1 = 3

t1 = 3/9.81

t1 = 0.3058 s

Now, for the time taken to achieve the distance (x1 + x2) on free-fall, we will use the formula;

s = ut + ½gt²

Where s = (x1 + x2) = 1.8 + 0.4587 = 2.2587 m

And now, u = 0 m/s because the start of the free fall is from maximum height with velocity of 0 m/s. Again, g = - 9.81 m/s²

Thus;

2.2587 = 0 - ½(-9.81)(t2)²

2.2587 = 4.905(t2)²

(t2)² = 2.2587/4.905

(t2)² = 0.4605

t2 = √0.4605

t2 = 0.6786 s

Thus, total time of feet in air = t1 + t2 = 0.3058 + 0.6786 = 0.9844 s

C) Velocity when feet hit the water would be given by;

v = u + gt

Where u = 0 m/s and t = t2 = 0.6786

Since it's in direction of gravity, g = 9.81 m/s

v = 0 + (0.6786 × 9.81)

v = 6.657 m/s

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Hannah walks 0.30 km to class in 5.0 min. what is her average speed in m/s?
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Answer:

1.0 m/s

Explanation:

First, convert to SI units.

0.30 km × (1000 m / km) = 300 m

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Speed is distance divided by time:

300 m / 300 s = 1.0 m/s

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A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p
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Answer:

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\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}

\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}

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and the tangential velocity of point B, and then the acceleration of point B:

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