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kati45 [8]
3 years ago
5

Inserting the formulas you found for Xman(t) and Xbus(t) into the conditionXman(tcatch)=Xbus(tcatch) , you obtain the following-

b+ct(catch) = 1/2 at^2( catch) or 1/2 at^2 (catch) - ct (catch) +b = 0Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man's speed (c) so that the equation above gives a solution for t catch that is a real positive number.Find Cmin the minimum value of c for which the man will catch the busExpress the minimum value for the man's speed in terms of a and b .
Physics
1 answer:
lara [203]3 years ago
3 0

Answer:

c > √(2ab)

Explanation:

In this exercise we are asked to find the condition for c in such a way that the results have been real

The given equation is

              ½ a t² - c t + b = 0

we can see that this is a quadratic equation whose solution is

             t = [c ±√(c² - 4 (½ a) b)]  / 2

for the results to be real, the square root must be real, so the radicand must be greater than zero

              c² -2a b > 0

              c > √(2ab)

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By holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. Therefore, the direction of the force on the charge you are holding will be to the southwest.

Let I hold the charge , q at the centre of given co-ordinate system and two positive charge of equal magnitude Q are placed 1 m to my North and 1 m to my South .

now, both the charge are same nature e.g., positive . Let my charge is also positive (well, you can assume negative too , I am considering positive because it makes me easy to solve) then, both charge repel to my charge.

charge Q placed on east is repelling my charge q toward west . similarly charge Q placed on North is repelling my charge q toward south.

Now , use vector for solve it.

vector F_{net} = vector Fe + vector Fn,

⇒ |F_{net}| = \sqrt{}  F^{2} _{e } + F^{2}_n

⇒ Fe = Fs = KqQ/(1m)² = KqQ

⇒ F_{net} = √{Fe² + Fs²} = √{(kqQ)²+(KqQ)²}

⇒ F_{net}= √2KqQ

Hence, net force act on q {my charge } is √2KqQ and the direction of force is S - W (southwest )direction.

To learn more about positive charges here

brainly.com/question/2903220

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