BITSAT2024PhysicsMotion in One DimensionActual
A particle is moving in a straight line. The variation of position ' x ' as a function of time ' t ' is given as x= (t^3-6 t^2+20 t+15 ) m . The velocity of the body when its acceleration becomes zero is:
Options
- A6 ~m / s
- B10 ~m / s
- C8 ~m / s
- D4 ~m / s
Correct answer
C. 8 ~m / s
Step-by-step solution
Displacement, x=t^3-6 t^2+20 t+15 Velocity, v= dx dt =3 t ^2-12 t +20 Acceleration, a= d v d t =6 t-12 When a=0 6 t -12=0 t=2 ~s At t =2 ~s , v=3(2)^2-12(2)+20=8 ~m / s