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3 years ago

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At 9:13AM a car is traveling at 35 miles per hour. Two minutes later, the car istraveling at 85 miles per hour. Prove that are s

ome time during this two-minuteinterval, the car’s acceleration is exactly 1500 miles per hour squared. Compute the value of A.

1 answer:

andreyandreev [35.5K]3 years ago

6 0

Answer:

There is at least one instant which instantaneous acceleration is equal to average acceleration. $a = 1500\,\frac{km}{h^{2}}$ .

Explanation:

The average acceleration experimented by the car is:

$\bar a = \frac{85\,\frac{km}{h} - 35\,\frac{km}{h} }{\frac{2}{60}\,h }$

$\bar a = 1500\,\frac{km}{h^{2}}$

According to the Rolle's Theorem, there is at least one instant t so that instantaneous acceleration equal to average acceleration for the analyzed interval. That is to say:

$v'(c) = \frac{v(\frac{2}{60} )-v(0)}{\frac{2}{60}-0}$

If car is accelerating at constant rate, instantaneous acceleration coincides with average acceleration for all instant t. Then, instantaneous acceleration is:

$a = 1500\,\frac{km}{h^{2}}$

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3. what could you do to increase the precision of your measurements? What factors limit the

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Use specific tools built specifically for that specific measurement.

Explanation:

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3 years ago

A 4.00-kg particle moves along the x axis. Its position varies with time according to x 5 t 1 2.0t 3, where x is in meters and t

White raven [17]

Answer:

Explanation:

Given equation is ,

x =t + 2 t³ ,

dx/dt = velocity ( v ) = 1 + 6 t²

a) kinetic energy = 1/2 m v² = .5 x 4 ( 1 + 6 t² )² = 2 ( 1 + 6 t²)²

b ) Acceleration = dv /dt = 12 t .

force( F ) = mass x acceleration = 4 x 12 t = 48 t

Power = force x velocity = 48 t x ( 1 + 6 t²). = 48 t + 288 t³ )

work done = ∫ F dx =∫ 48 t x( 1 + 6t² )dt ; = [48t²/2 + 48 x 6 x t³ /3 = 24 t² + 96 t³ )]₀² = 864 J

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2 years ago

Read 2 more answers

An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F

ololo11 [35]

Answer

given,

mass of jogger = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed = 1.3 m/s in 61 ° north of east.

jogger one

$P_1 = m_1 v_1 \hat{i}$

$P_1 = 67 \times 2.3\hat{i}$

$P_1 = 154.1 \hat{i}$

$P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}$

$P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}$

$P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}$

$P_2 = 44.12\hat{i} +79.59\hat{j}$

now

P = P₁ + P₂

$P = 198.22 \hat{i} +79.59 \hat{j}$

magnitude

$P = \sqrt{198.22^2 + 79.59^2}$

$P =213.60 kg.m/s$

$\theta = tan^{-1}\dfrac{79.59}{198.22}$

$\theta = 21.87$

the angle is $\theta = 21.87$ north of east

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3 years ago

Suppose you push a hockey puck of mass m across frictionless ice for a time \Delta t, starting from rest, giving thepuck speed v

bazaltina [42]

Answer:

1. $t_2 = 2t_1$

2. $t_2 = t_1\sqrt{2}$

Explanation:

1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.

In Newton's 2nd law: F = ma

where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.

$a_1 = 2a_2$

Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time

$t = v / a_1$

Now that acceleration is halved:

$t = \frac{v}{2a_2}$

$\frac{v}{a_2} = 2t$

You would need to push for twice amount of time $t_2 = 2t_1$

2. The distance traveled by the puck is as the following equation:

$d = at^2$

So if the acceleration is halved while maintaining the same d:

$\frac{d_1}{d_2} = \frac{a_1t_1^2/2}{a_2t_2^2/2}$

As $d_1 = d_2$ , then $d_1/d_2 = 1$ . Also $a_1 = 2a_2$

$1 = \frac{2a_2t_1^2}{a_2t_2^2}$

$t_2^2 = 2t_1^2$

$t_2 = t_1\sqrt{2}\approx 1.14t_1$

So t increased by 1.14

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3 years ago

Each lunar cycle has one full moon, in which the relative positions of Earth, the sun, and the moon form in a straight line

aniked [119]

Complete Question:

Each lunar cycle has one full moon, in which the relative positions of Earth, the sun, and the moon form in a straight line. Which list represents the position of Earth, the sun, and the moon during a full moon?

Group of answer choices.

A. Earth, sun, moon

B. sun, moon, Earth

C. moon, sun, Earth

D. sun, Earth, moon

Answer:

D. sun, Earth, moon

Explanation:

A lunar eclipse is a phenomenon that occurs when the Earth comes between the Moon and the Sun thereby causing it to cover the Moon with its shadow.

Simply stated, lunar eclipse takes place when the Moon passes or moves through the Earth's shadow thereby blocking any ray of sunlight from reaching the Moon. Thus, the full moon appears deep red (blood moon).

Also, a lunar eclipse would occur only when the Sun, Earth, and Moon are closely aligned to form a straight line known as the syzygy.

There are three (3) types of lunar eclipse and these are;

1. Total lunar eclipse.

2. Partial lunar eclipse.

3. Penumbra lunar eclipse.

Each lunar cycle has one full moon, in which the relative positions of Earth, the sun, and the moon form in a straight line. Thus, the list which represents the position of Earth, the sun, and the moon during a full moon is sun, Earth, and moon

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3 years ago