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The velocity of a particle traveling in a straight line is given by v=(6t−3t2) m/s, where t is in seconds. Suppose that s = 0 when t = 0.

Part A: Determine the particle's deceleration when t = 3.1 s

Part B: Determine the particle's position when t = 3.1 s .

Part C: How far has the particle traveled during the 3.1-s time interval?

Part D: What is the average speed of the particle for the time period given in previous part?

Respuesta :

MechEngineer

Answer:

A) 12.6 m/s2

B) -0.961 m

C) 0.961 m

D) 0.31 m/s

Explanation:

We can derive the acceleration function in term of t by taking 1st derivative of v(t)

[tex]a(t) = \frac{dv(t)}{dt} = 6 - 6t[/tex]

The position function in term of t would be the integration of v(t)

[tex]s(t) = \int {v(t)} \, dt=\int {(6t - 3t^2)} \, dt = 3t^2 - t^3 + s_0[/tex]

Since s = 0 when t = 0 we can conclude that [tex]s_0 = 0[/tex] by plugging in t = 0 and s = 0

Part A: a(t = 3.1) = 6 - 6*3.1 = -12.6 m/s2. So the particle's deceleration is 12.6 m/s2 when t = 3.1

Part B:

[tex]s(t = 3.1) = 3*3.1^2 - 3.1^3 = -0.961 m [/tex]

Part C: As s(0) = 0 and s(3.1) = -0.961. The particle has traveled a distance of |-0.961 - 0| = 0.961m

Part D: So the particle has traveled a distance of 0.961m within time of 3.1s. That means the average speed is overall distance divided by overall time

= 0.961 / 3.1 = 0.31m/s

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