AP EAMCET201920 Apr 2019Morning ShiftPhysicsMotion in One DimensionActual
The velocity of an object moving in a straight line path is given as a function of time by (v=6 t-3 t^2 ), where (v ) is in ( ms ⁻¹, t ) is in ( s ). The average velocity of the object between, (t=0 ) and (t=2 ~s ) is
Options
- A0
- B(3 ~ms ⁻¹ )
- C(2 ~ms ⁻¹ )
- D(4 ~ms ⁻¹ )
Correct answer
C. (2 ~ms ⁻¹ )
Step-by-step solution
Given, velocity, (v=6 t-3 t^2 ) As we know that, (v= d x d t ) Here, (x ) is the displacement of the particle. Now, (d x=v d t ) Integrate on the both sides, limit (t=0 ) to (t=2 ), we get ( aligned x & = ₀^2 v d t= ₀^2 (6 t-3 t^2 ) d t & = [ 6 t^2 2 ]₀^2- [ 3 t^3 3 ]₀^2= [3 t^2 ]₀^2- [t^3 ]₀^2 & = [3(2)^2-3(0)^2 ]- [(2)^3-(0)^2 ] & =[12-0]-[8-0]=12-8=4 ~m aligned ) ( aligned Average velocity, v_ avg & = Total displacement Total time taken & = 4 2 =2 ~m / s aligned ) Hence, the average velocity of the object betwee