Part B ,C are wrong plz help. Problem 12.9 Th e acceleration of a particle as it
ID: 1717672 • Letter: P
Question
Part B ,C are wrong plz help. Problem 12.9 Th e acceleration of a particle as it moves along a straight line is given by a (2t-1) m/s2, where t is in seconds. Suppose that s = 6 m and u 7 m/s when t 0. Part A Determine the particle's velocity when t = 4 s . Express your answer to three significant figures and include the appropriate units. u= 190 Submit Give Up Correct Part B Determine the particle's position when t = 4 s Express your answer to three significant figures and include the appropriate units. s= 27.33 |mExplanation / Answer
>> As, V = Velocity
As, acceleration, a = dv/dt
=> a = (2t - 1)m/s2
As, a = dv/dt
=> dv = a dt
=> dv = (2t - 1) dt
Integrating both sides
=> v = t2 - t + C [ C = Integration Constant ]
Now at t = 0, v = 7 m/s
=> C = 7
=> V = t2 - t + 7
>> As, s = Displacement and,
v = ds/dt
=> ds = v dt
=> ds = (t2 - t + 7 ) dt
Integrating both sides,
=> s = (1/3)t3 - (1/2)t2 + 7t + D [ D = Constant of Integration ]
>> As, at t = 0, s = 6 ( Given)
=> D = 6
=> s = s = (1/3)t3 - 0.5t2 + 7t + 6
So, at t= 4 sec,
particle's position, s = (1/3)*4*4*4 - 0.5*4*4 + 7*4 + 6 = 47.33 m
=> Particle's Position at t = 4 sec = 47.33 m ...answer...part (2)
>> As, assume at some time t", particle has zero velocity and it started moving backwards.
As, v = t2 - t + 7
=> 0 = (t")2 - t" + 7
=> As, v is not zero for any real value of t
So, our particle does not move backwards
=> So, total displacement = total distance
=> Total distance moved during the 4 seconds = 47.33 m ...ANSWER....PART (3)...