0 points SerPSE9 2.P.020 W An object moves along the x axis according to the equ

Question

0 points SerPSE9 2.P.020 W An object moves along the x axis according to the equation-2.752-2.00t+3.00, where x is in meters and t is in seconds. a) Determine the average speed between t-3.50 s and t 4.70s m/s (b) Determine the instantaneous speed at t 3.50 s. m/s Determine the instantaneous speed at t = 470 s. m/s (c) Determine the average acceleration between t 3.so s and r = 4.70 s. (d) Determine the instantaneous acceleration at t 3.50s m/s Determine the instantaneous acceleration at t 4.70 s. m/s (e) At what time is the object at rest? Need Help? Read it

Explanation / Answer

x =2.75 t^2 - 2 t + 3

We Know,

Velocity = distance / time

v = dx/dt

v = d/dt (2.75 t^2 - 2 t + 3)

v = 2* 2.75 t - 2

v = 5.5t -2 m/s

Acceleration = Velocity /Time

a = dv/dt

a = d/dt(5.5t -2)

a = 5.5 m/s^2

(a)

Average Speed between t =3.5 and t = 4.7

Position x @ t = 3.5

x =2.75 (3.5)^2 - 2*3.5 + 3

x = 29.69 m

Position x @ t = 4.7

x =2.75 (4.7)^2 - 2*4.7 + 3

x = 54.35 m

Average speed = xf -xi / tf-fi

Average speed = 54.35 - 29.69 / 4.7 -3.5

Average speed = 22.20 / 1.2

Average speed = 20.55 m/s

b)

Instantaneous Speed at t = 3.5s

v = 5.5*3.5 -2

v = 17.25 m/s

c)

Instantaneous Speed at t = 4.7s

v = 5.5*4.7 -2
v = 23.85 m/s

d)

Average acceleration between t =3.5 and t = 4.7

a = vf - vi / tf-ti

a = 23.85 - 17.25 / 4.7 -3.5

Average acceleration a = 5.5 m/s^2

e)
As acceleration is constant-
Instantaneous acceleration at t = 3.5s

a = 5.5 m/s^2

f)
As acceleration is constant-
Instantaneous acceleration at t = 4.7s
a = 5.5 m/s^2

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