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Tafu is working with subatomic particles in the Physics lab. A positron is traveling in a straight line down the particle accelerator. Tafu models its position with the function
p(t)= t^2/t – 1

(g) Find a time t>1 when the positron's instantaneous velocity is zero
(h) Where is the positron located at the time you found in (g)

Respuesta :

Answer:

g) 2 h) 4

Step-by-step explanation:

g) The time t when the instantaneous velocity of the positron is zero is;

t = 2

h) The location of the positron at the time of t = 2 is;

p = 4.

g) We are told that the function that models the position of the sub atomic particles is;

p(t) = t²/(t - 1)

Instantaneous velocity is simply the derivative of the distance with respect to time.

Thus, Instantaneous velocity is gotten by using online differentiation calculator to get; dp/dt = [2t/(t - 1)] - [t²/(t - 1)²]

When Instantaneous velocity is zero, we have;

[2t/(t - 1)] - [t²/(t - 1)²] = 0

Rearranging to get;

[2t/(t - 1)] = [t²/(t - 1)²]

Cross multiply to get;

2t(t - 1)² = t²(t - 1)

divide both sides by t to get;

2(t - 1)² = t(t - 1)

Expand the bracket to get;

2(t² - 2t + 1) = t² - t

2t² - 4t + 2 = t² - t

Rearrange to get;

t² - 3t + 2 = 0

Using online quadratic equation solver gives;

t = 1 or 2

We are told that t > 1

Thus, we will adopt t = 2

h) At t = 2 seconds gotten in g above, the positron is located at;

p(2) = 2²/(2 - 1)

p(2) = 4

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