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antoniya [11.8K]

3 years ago

13

Two cars, initially at rest and 5 km apart at t=0 , simultaneously move toward each other. Car A travels at a constant speed of

55 km/hour, while car B travels at a constant speed of 12 km/hour. How long will it take care from t=0 for car A to meet car B?
Select one:
a. 24.5 s
b. 490 s
c. 245 s
d. 268 s
e. None of the above.

1 answer:

Anastasy [175]3 years ago

8 0

Answer:

<em>d. 268 s</em>

Explanation:

<u>Constant Speed Motion</u>

An object is said to travel at constant speed if the ratio of the distance traveled by the time taken is constant.

Expressed in a simple equation, we have:

$\displaystyle v=\frac{d}{t}$

Where

v = Speed of the object

d = Distance traveled

t = Time taken to travel d.

From the equation above, we can solve for d:

d = v . t

And we can also solve it for t:

$\displaystyle t=\frac{d}{v}$

Two cars are initially separated by 5 km are approaching each other at relative speeds of 55 km/h and 12 km/h respectively. The total speed at which they are approaching is 55+12 = 67 km/h.

The time it will take for them to meet is:

$\displaystyle t=\frac{5}{67}$

t = 0.0746 hours

Converting to seconds: 0.0746*3600 = 268.56

The closest answer is d. 268 s

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(b): $\rm meter/ second^3.$

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Explanation:

Given, the position of the particle along the x axis is

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The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

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(a):

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(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

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Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

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Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

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Answer:

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