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Kisachek [45]
3 years ago
15

Traveler A starts from rest at a constant acceleration of 6 m/s^2. Two seconds later, traveler B starts with an initial velocity

of 20 m/s at the same acceleration of 6 m/s^2. as measured by a, at what time will traveler B overtake traveler A?
a. 0.4s
b. 1.5s
c. 2.0s
d. 2.5s
e. 3.5s
Physics
1 answer:
Troyanec [42]3 years ago
4 0

Answer:

3. 3.5 s

Explanation:

The position of traveller A is given by the equation:

x_A(t) = \frac{1}{2}a t^2

where

a = 6 m/s^2 is the acceleration of A

t is the time measured from when A started the motion

The position of traveller B instead is given by

x_B(t) = u_B (t-2) + \frac{1}{2}a(t-2)^2

where a (acceleration) is the same as traveller A, and

u_B = 20 m/s

is B's initial velocity. We can verify that the formula is correct by substituting t=2, and we get x_B=0, which means that B starts its motion 2 seconds later.

Traveller B overtakes traveller A when the two positions are the same, so:

x_A = x_B\\\frac{1}{2}at^2 = u_B (t-2) + \frac{1}{2}a(t-2)^2\\\frac{1}{2}at^2 = u_B t - 2u_B +\frac{1}{2}at^2 +2a-2at\\u_Bt-2at = 2u_B-2a\\t=\frac{2u_B-2a}{u_B-2a}=\frac{2(20)-2(6)}{20-2(6)}=3.5 s

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What conversion factor is used to convert 20 cm to m
Anastaziya [24]
The conversion factor you use is 100 cm = 1 m.
You can divide 20 by 100 to get the answer.
20 cm/100 cm =.2 m
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3 years ago
A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate
Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by b) is at 36 ft. from the wall is the following:

\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s

Explanation:

The Area of the triangle is given by A=h\times b where h=\sqrt{l^2-b^2} (by using the Pythagoras' Theorem) and b is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

A=\sqrt{l^2-b^2}b

The rate of change of the area is given by its time derivative

\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)

\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}

\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Product rule

\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Chain rule

\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}

\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)

In here we can identify b=36\, ft, l=39 and \frac{db}{dt}=8\,ft/s.

The result is then

\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s

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The fig below explains it.

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