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2 years ago

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You run due east at a constant speed of 3.00 m/s for a distance of 120.0 m and then continue running east at a constant speed of

5.00 m/s for another 120.0 m. For the total 240.0-m run, is your average velocity 4.00 m/s, greater than 4.00 m/s, or less than 4.00 m/s? Explain.

1 answer:

Leni [432]2 years ago

7 0

Answer:

Explanation:

Given

Speed while running towards east is $v_1=3\ m/s$

Distance traveled in east direction $x_1=120\ m$

For Another interval you run with velocity

$v_2=5\ m/s$

$x_2=240\ m$

Total displacement $=x_1+x_2$

$=120+120=240\ m$

Time for first interval

$t_1=\frac{x_1}{v_1}=\frac{120}{3}$

$t_1=\frac{120}{3}=40\ s$

Time for second interval

$t_2=\frac{x_2}{v_2}=\frac{120}{5}=24\ s$

total time $t=t_1+t_2$

$t=40+24=64\ s$

average velocity $v_{avg}=\frac{x_1+x_2}{t}$

$v_{avg}=\frac{240}{64}=3.75\ m/s$

Therefore average velocity is less than $4 m/s$

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A displacement vector points in a direction of θ = 23° left of the positive y-axis. The magnitude of this vector is D = 155 m. R

Lady bird [3.3K]

Answer:

Dₓ = -155 sin 23° i + 0 j

Explanation:

The diagram showing the vector has been attached to this response.

As shown in the diagram,

The vector D has an x-component (also called horizontal component) of -D sinθ i. i.e

Dₓ = -D sin θ i [The negative sign shows that D lies in the negative x direction]

Where;

D = magnitude of D = 155m

θ = direction of D = 23°

Therefore;

Dₓ = -155 sin 23° i

Since Dₓ represents the x component, its unit vector, j component has a value of 0.

Therefore, Dₓ can be written in terms of D, θ and the unit vectors i and j as follows;

Dₓ = -155 sin 23° i + 0 j

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2 years ago

A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by

aleksklad [387]

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The position of the car as a function of time t is given by

$x(t)=\alpha t^2 - \beta t^3$

where

$\alpha = 1.50 m/s^2$

$\beta = 0.05 m/s^3$

The average velocity is given by the ratio between the displacement and the time taken:

$v=\frac{\Delta x}{\Delta t}$

The position at t = 0 is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

The position at t = 2.00 s is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

So the displacement is

$\Delta x = x(2)-x(0)=5.6-0=5.6 m$

The time interval is

$\Delta t = 2.0 s - 0 s = 2.0 s$

And so, the average velocity in this interval is

$v=\frac{5.6 m}{2.0 s}=2.8 m/s$

b)

The position at t = 0 is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

While the position at t = 4.00 s is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

So the displacement is

$\Delta x = x(4)-x(0)=20.8-0=20.8 m$

The time interval is

$\Delta t = 4.0 - 0 = 4.0 s$

So the average velocity here is

$v=\frac{20.8}{4.0}=5.2 m/s$

c)

The position at t = 2 s is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

While the position at t = 4 s is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

So the displacement is

$\Delta x = 20.8 - 5.6 = 15.2 m$

While the time interval is

$\Delta t = 4.0 - 2.0 = 2.0 s$

So the average velocity is

$v=\frac{15.2}{2.0}=7.6 m/s$

Learn more about average velocity:

brainly.com/question/8893949

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Answer:

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Answer:

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