Ask question

Ask question

All categories

2 years ago

13

Which times and distances are represented by the function? select three options. the starting distance, at 0 hours, is 300 miles

. at 2 hours, the traveler is 725 miles from home. at 2.5 hours, the traveler is still moving farther from home. at 3 hours, the distance is constant, at 875 miles. the total distance from home after 6 hours is 1,062.5 miles.

1 answer:

Naya [18.7K]2 years ago

4 0

The times and distances are represented by the function are

at 2 hours, the traveler is 725 miles from home.
at 3 hours, the distance is constant, at 875 miles
the total distance from home after 6 hours is 1,062.5 miles.

<h3>What is distance?</h3>

The distance is the length of path travelled by a body when moving with some speed and taking time t.

Given function is

D (t) = 300t +125 for 0≤t<2.5

D (t) = 875 for 2.5 ≤t≤3.5

D(t)= 75t +612.5 for 3.5<t≤6

According to the given function, for 2 hours, the traveler will travel from home,

D (2) = 300x2 +125 = 725 miles.

For time between 2.5 to 3.5 hr, the distance is constant i.e. 875 miles.

For total time 6hr, the distance travelled will be

D(6) = 75 x 6 +612.5 = 1062.5 miles.

Thus, the correct options are picked up.

Learn more about distance.

brainly.com/question/15172156

#SPJ1

You might be interested in

If it takes Ashley 6.1 seconds to run at an average speed of 5.7 meters per second, what is the distance (in meters) she covers

Softa [21]

Answer:

34.8m

Explanation:

distance = speed x time

6.1 x 5.7 = 34.77

34.8

3 0

3 years ago

A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w

cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

$\displaystyle P=\frac{W}{t}$

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

$W=F.x$

The second Newton's law gives us the net force as

$F=m.a$

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

$v_f=v_o+at$

$v_f=at$

Solving for a

$\displaystyle a=\frac{v_f}{t}={5.4}{1}$

$a=5.4\ m/s^2$

The distance traveled in the interval is given by

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

Since vo=0

$\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}$

$x=2.7\ m$

The force is given by

$F=m.a$

We don't know the value of m, so the force is

$F=2.7m$

Computing the work done by the sprinter

$W=F.x=2.7m(5.4)$

$W=14.58m$

The power is finally computed

$\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}$

$P=14.58m$

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

$\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}$

$a=8.8\ m/s^2$

The distance is

$\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}$

$x=3.8\ m$

The net force is

$F=m(8.8)=8.8m$

The work done by the sprinter is now computed as

$W=8.8m(3.8)=33.44m$

At last, the output power is

$\displaystyle P=\frac{33.44m}{0.5}=66.88m$

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0

4 years ago

What is the relationship between force and motion described by Newton's first law

prisoha [69]

Newtons First Law of Motion:
An object at rest stays at rest and an object in motion<span> stays in </span>motion <span>with the same speed and in the same direction unless acted upon by an unbalanced force.</span>

Therefore, the relationship between force and motion is that it takes force to change the speed or direction of any object in motion.

4 0

3 years ago

Help with these please someone?

DIA [1.3K]

Answer:

8. 2.75·10^-4 s^-1

9. No, too much of the carbon-14 would have decayed for radiation to be detected.

Explanation:

8. The half-life of 42 minutes is 2520 seconds, so you have ...

1/2 = e^(-λt) = e^(-(2520 s)λ)

ln(1/2) = -(2520 s)λ

-ln(1/2)/(2520 s) = λ ≈ 2.75×10^-4 s^-1

___

9. Reference material on carbon-14 dating suggests the method is not useful for time periods greater than about 50,000 years. The half-life of C-14 is about 5730 years, so at 65 million years, about ...

6.5·10^7/5.73·10^3 ≈ 11344

half-lives will have passed. Whatever carbon 14 may have existed at the time will have decayed completely to nothing after that many half-lives.

7 0

3 years ago

Canadian geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at

Wewaii [24]

Answer:

θ=19.877⁰

Explanation:

Given data

Velocity Va=34.0 km/h

Velocity Va=100 km/h

To find

Angle θ

Solution

We want the bird to fly with velocity Vb=100 km/h with an angle θ relative to the ground so that the bird fly due south relative to the ground.From figure which is attached we got

Sinθ=(Va/Vb)

Sinθ=(34.0/100)

θ=Sin⁻¹(34.0/100)

θ=19.877⁰

5 0

3 years ago