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Annette [7]
4 years ago
9

An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west. It start

s at the 16 inch mark, crawls to the 29-inch mark, then moves to the 14-inch mark. What is the total distance the ant traveled? What was the total displacement of the ant?
Physics
1 answer:
FrozenT [24]4 years ago
6 0

Answer:

  • The total distance traveled is 28 inches.
  • The displacement is 2 inches to the east.

Explanation:

Lets put a frame of reference in the problem. Starting the frame of reference at the point with the 0-inch mark, and making the unit vector \hat{i} pointing in the west direction, the ant start at position

\vec{r}_0 = 16 \ inch \ \hat{i}

Then, moves to

\vec{r}_1 = 29 \ inch \ \hat{i}

so, the distance traveled here is

d_1 = |\vec{r}_1 - \vec{r}_0  | = | 29 \ inch   \ \hat{i} - 16 \ inch   \ \hat{i}  |

d_1 =  | 13 \ inch   \ \hat{i}  |

d_1 =  13 \ inch

after this, the ant travels to

\vec{r}_2 = 14 \ inch \ \hat{i}

so, the distance traveled here is

d_2 = |\vec{r}_2 - \vec{r}_1  | = | 14 \ inch   \ \hat{i} - 29 \ inch   \ \hat{i}  |

d_2 =  | - 15 \ inch   \ \hat{i}  |

d_2 =  15 \ inch

The total distance traveled will be:

d_1 + d_2 = 13 \ inch + 15 \ inch = 28 \ inch

The displacement is the final position vector minus the initial position vector:

\vec{D}=\vec{r}_2 - \vec{r}_1

\vec{D}= 14 \ inch   \ \hat{i} - 16 \ inch \ \hat{i}

\vec{D}= - 2 \ inch \ \hat{i}

This is 2 inches to the east.

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Explanation:

We have,

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It is required to find which plane is faster. To find which plane is faster, we need to compare their speeds.

Speed of a plane a is 900 km/h and that of plane b is 50 km/h. So, we can say that plane a is moving faster.

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A diver jumps off a cliff 50m high and needs to clear the rock that extend outward 5.0m from the base of the cliff. The diver ju
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Answer:

He should run at least at 1.5 m/s

The diver will enter the water at an angle of 87° below the horizontal.

Explanation:

Hi there!

The position and velocity of the diver are given by the following vectors:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

v = (v0x, v0y + g · t)

Where:

r = position vector at time t

x0 = initial horizontal position

v0x = initial horizontal velocity

t = time

y0 = initial vertical position

v0y = initial vertical velocity

g = acceleration due to gravity (-9.8 m/s² considering the  upward direction as positive)

v = velocity vector at time t

Please, see the attached figure for a description of the problem. Notice that the origin of the frame of reference is located at the jumping point so that x0 and y0 = 0.

We know that, to clear the rocks, the position vector r final (see figure) should be:

r final = ( > 5.0 m, -50 m)

So let´s find first at which time the y-component of the vector r final is - 50 m:

y = y0 + v0y · t + 1/2 · g · t²

-50 m = 2.1 m/s · t - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 2.1 m/s · t + 50 m

Solving the quadratic equation

t = 3.4 s

Now, we can calculate the initial horizontal velocity using the equation of the x-component of the position vector knowing that at t =3.4 the horizontal component should be greater than 5.0 m:

x = x0 + v0x · t      (x0 = 0)

5.0 m < v0x · 3.4 s

v0x > 5.0 m / 3.4 s

v0x > 1.5 m/s

The initial horizontal velocity should be greater than 1.5 m/s

To find the angle at which the diver enters the water, we have to find the magnitude of the final velocity (vector vf in the figure). We already know the magnitude of the x-component of the vector vf, since the horizontal velocity is constant. So:

vfx > 1.5 m/s

Now, let´s calculate vfy:

vfy = v0y + g · t

vfy = 2.1 m/s - 9.8 m/s² · 3.4 s

vfy = -31 m/s

Let´s calculate the minimum magnitude that the final velocity will have if the diver safely clears the rocks. Let´s consider the smallest value allowed for vfx: 1.5 m/s. Then:

|v| = \sqrt{(1.5 m/s)^{2} + (31m/s)^{2}} = 31 m/s

Then the final velocity of the diver will be greater or equal than 31 m/s.

To find the angle, we have to use trigonometry. Notice in the figure that the vectors vf, vfx and vy form a right triangle in which vf is the hypotenuse, vfx is the adjacent side and vfy is the opposite side to the angle. Then:

cos θ = adjacent / hypotenuse = vfx / vf = 1.5 m/s / 31 m/s

θ = 87°

The diver will enter the water at an angle of 87° below the horizontal.

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The relevant equation to use here is:

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Explanation:

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