NTA Abhyas JEE Main2020PhysicsMotion in One DimensionPractice
A particle moves along a straight line with a variable acceleration given in the acceleration-displacement a - S curve as shown in the figure. Determine the velocity (in m   s - 1 ) of the particle after it has travelled a distance of 30   m . The initial velocity is 10   m  s - 1 .
Correct answer
20
Step-by-step solution
Area under curve is = 1 2 × 10 × 30 = 150 ……..(i) Area under the curve is also equal to = v 2 - u 2 2 …….(ii) From (i) and (ii) 1 2 v 2 - u 2 = 150 v 2 - u 2 =300 v 2 = u 2 + 300 v 2 = 10 2 + 300 v = 400 = 20 m s - 1