Car A is traveling at a constant velocity of upsilon_A in the + x-direction. At

Question

Car A is traveling at a constant velocity of upsilon_A in the + x-direction. At the moment when car A is at x = d along the positive x-axis, car B begins at x = 0m and accelerates from rest with an acceleration of a_B and catches up to car A after a time t. Find a_B. (d - upsilon_A t)/(2t^2) B) 2 (d - upsilon_A t)/(t^2) C) 2 (d + upsilon_a t)/(t^2) D) (d + upsilon_a t)/(2t^2) Two objects are rotating about the same axis. Object A has a moment of inertia of 4kg middot m^2 and object B has a moment of inertia of 12kg m^2. Object A is initially rotating with an angular frequency of omega_A = 2rad/s and object B is initially rotating with an angular frequency of -4rad/s. The two objects are linked together without the aid of any net external torque so that they rotate as a single unit with a common angular frequency, omega, about the same axis as before. What is rotate as a single unit with a common angular frequency, omega, about the same axis as before. What is omega? omega = - 3.5rad/s B) omega = - 3.0rad/s C) omega = - 2.5rad/s D) omega = - 2.0rad/s E) omega = - 1.5rad/s F) omega = - 1.0rad/s

Explanation / Answer

Constant velocity of the car A= vA

Initial velocity of car B, uB= 0 (since it starts from rest)

Acceleration of car B= aB

Initial distance between both cars when car B starts= d

Both cars meet after time t. Then using equation of motion, distance covered by car B in time "t",

dB= uBt+(1/2)aBt2

dB= 0+(1/2)aBt2............................(1)

Since acceleration of car A is 0 hence distance covered by it in time "t" is

dA= vAt+0= vAt...........................(2)

Then from the given scenario, we can write,

dB= d+dA

using equation 1 and 2 in above,

  (1/2)aBt2= d+vAt

aB= 2(d+vAt)/t2   (ANS)

Hence correct option: C

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---