an object moves along the x axis according to the equation x=2.80t*2 -2.00t ÷3.0

Question

an object moves along the x axis according to the equation x=2.80t*2 -2.00t ÷3.00, where x is in meters and t is in seconds.(a) determine the average speed between t = 2.20s and t= 3.80s in m/s. (b) determine the instantaneous speed at 2.20s in m/s. determine the instantaneous speed at t= 3.80s in m/s. (c) determine the average acceleration between t= 2.20 s and t= 3.80s in m/s*2(d) determine the instantaneous acceleration at t= 2.20s in m/s*2. determine the instantaneous acceleration at t= 3.80s in m/s*2. (e) at what time is the objectat rest?

Explanation / Answer

here,

x = (2.80t^2 -2.00t)/3

dx/dt = d' = (5.6t - 2)/3

d'' = 5.63/3 = 1.877

at t = 2.20 , d'(t1) = (5.6*2.2 -2)/3 = 3.44
at t = 3.80 , d'(t2) = (5.6*3.8 - 2)/3 = 6.427

Part A:
Avg speed = net distance /total time
Savg = ( x(t2) - x(t1) ) /(t2 - t1)
Savg = ( ( (2.80(3.8)^2 -2.00*3.80)/3 ) - ( (2.80*(2.2)^2 -2.00*2.2)/3 ) ) / (3.80 - 2.2)
Savg = 4.933 m/s or 5 m/s(rounded off)

part B:
Inst velocity at t = 2.20s = 3.44 m/s
Inst velocity at t = 3.80s = 6.427 m/s

Part C:
average acc = (v2 -v1)/ (t2-t1)
Aavg = ( d'(t2) - d'(t1) ) / (t2 - t1)
Aavg = ( 6.427 - 3.44 ) / (3.80 - 2.2)
Aavg = 1.867 m/s^2 or 2 m/s^2(rounded off)

Part D:
Instantenous acceleration will be same for both time periods i.e = d'' = 1.877 m/s^2

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