9:30 PM webassign.net T-Mobile Wi-Fi Find the instantaneous velocity of the part

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9:30 PM webassign.net T-Mobile Wi-Fi Find the instantaneous velocity of the particle described in the figure below at the following times. (a) t=1.1s The instantaneous velocity at a particular time is the slope of the position versus time graph at that instant in time. m/s (b) t = 3.1 s The instantaneous velocity at a particular time is the slope of the position versus time graph at that instant in time. m/s (c) t = 4.8 s What is the displacement of the particle between 4 and 5 seconds? m/s (d) t = 7.6 s How does the slope of the line between 7 and 8 s compare to the slope in the first 2 seconds? m/s Need Help? osdheach An object moves along the x axis according to the equation 2-200c + 3.00. wherex is in meters and t is in seconds. (a) Determine the average speed between t 2.90 s and t = 4.40 s. (b) Determine the instantaneous speed at t = 2.90 s. Determine the instantaneous speed at t-4.40 s. We are given an explicit expression for x as a function of time. Calculus methods can be used to determine the velocity as a function of time m/s 4.40 (c) Determine the average acceleration between t-2.90 s and t 5. (d) Determine the instantaneous acceleration at t 2.90 s. m/s Determine the instantaneous acceleration at t 4.40 s. m/s (e) At what time is the object at rest?

Explanation / Answer

1)

a)

Instantaneous velocity , v = 10/2 =5 m/s

b)

v = -5/4 = -1.25 m/s

c)

Instantaneous velocity, v = 0 <----- as the slope of the curve at this moment is 0

d)

v = 5/1 = 5 m/s

2)

c)

x = 2.9*t^2 - 2*t +3

So, v = dx/dt = 2.9*2*t - 2 = 5.8*t - 2

So, average acceleration = 5.8*(4.4 - 2.9)/(4.4 - 2.9) = 5.8 m/s2

d)

a = dv/dt = 5.8 m/s2 <------ constant

So, instantaneous accelertion (at t = 2.9 )= 5.8 m/s2.

Similalrly, the instantaneous acceleration ( at t = 4.4 s) = 5.8 m/s2

e)

v = 5.8*t - 2 = 0

Si=o, t = 2/5.8 = 0.344 s

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