The velocity of the center of mass of a system of particles changes as v = 4.5 t

Question

The velocity of the center of mass of a system of particles changes as v = 4.5 t + 2.4 t^2 + 11 t^3. where v is in meters per second. If the system starts from rest at t = 0, what is its acceleration at t = 3.0 s? 7.1 m/s^2 14 m/s^2 25 m/s^2 49 m/s^2 65 m/s^2 Momentum may be expressed in: kg/m gram s N s kg (m s) N/s A 1.0 kg-ball moving at 2.0 m's perpendicular to a wall rebounds from the wall at 1.5 m/s. The change in the momentum of the ball is: zero 0.5 N-s away from wall 0.5 N s toward wall 3.5 N s away from wall 3.5 N s toward wall A 5-kg object can move along the x axis. It is subjected to a force in the positive x direction; a graph of F as a function of time t is shown below. Over the time the force is applied the change in the velocity of the object is: 0.8 m/s Urn's 1.6 m/s 2.3 m/s 4.0 m/s One revolution per minute is about: 0.0524 rad/s 0.105 rad/s 0.95 rad/s 1.57 rad/s 6 28 rad/s

Explanation / Answer

6.

v = 4.5*t + 2.4*t^2 + 1.1*t^3

acceleration is given by

a = dv/dt

a = d(4.5*t + 2.4*t^2 + 1.1*t^3)/dt

a = 4.5 + 2*2.4*t + 3*1.1*t^2

a = 4.5 + 4.8*t + 3.3*t^2

at t = 3.0 sec

a = 4.5 + 4.8*3 + 3.3*3.0^2

a = 48.6 m/sec^2 = 49 m/sec^2

7.

momentum = mass*velocity = force*t

mass = kg

velocity = m/sec

momentum = kg-m/sec

force = N = kg-m/sec^2

time = sec.

momentum = time*force

momentum = N*s

8.

change in momentum = Pi - Pf

Pi = 1*2 = 2 kg-m/sec

Pf = 1*(-1.5) = -1.5 kg-m/sec

dP = 2 - (-1.5) = 3.5 kg-m/sec towards the wall

9.

P = m*dv = F*dt

5*dv = 4*(3 - 1)

dV = 8/5 = 1.6 m/sec

10.

1 rev/min = 0.105 rad/sec

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