4 At t0 Block A starts from rest and moves down to the left

4. At t-0. Block \"A\" starts from rest and moves down to the left with a constant acceleration of 48 mm/s Determine a) the acceleration of block \"B\" b) the velocity of block \"B\" relative to block \"A \" when 5s

Solution

given,

Constant Acceleration of Block A, a_A = 48 mm/s² = 0.048 m/s²

Also Initial velocity of Block A and Block B , u_A = u_B= 0 m/s (Initially at rest at t=0s)

Suppose,

Accleration of Block B = a_B

Velocity of Block A = v_{A at t= 5s}

Velocity of Block B = v_{B at t =5s}

Distance moved by the Block A = S_{A in t=5}

Distance moved by the Block B = S_{B in t=5}

Now using Kinematic equation for constant acceleration for Block A,

S_A = u_At + 1/2*a_A*t²

S_A = 1/2*0.048*25 (u_A= 0)

= 0.6m

v_A² = u_A² + 2*a_A*S_A

= 0 + 2*0.048*0.6 = 0.0576

v_A = 0.24 m/s

Also from figure it is very clear that,

a_A = 2 a_B

v_A = 2 v_B

_therefore

Acceleration of Block B, a_B = 0.048/2 = 0.024 m/s²

v_B = 0.24/2 = 0.12 m/s

Velocity of Block B with respect to A

V_{BA at t}= 5s = vB - vA = 0.12 - 0.24 = -0.12 m/s (-ve sign shows that velocity of Block B is Opposite to that of Velocity of Block A).