JEE Main Physics question
Given that the position of the body in metre is a function of time as follows: x = 2t⁴ + 5t + 4. The mass of the body is 2 kg. What is the increase in its kinetic energy one second after the start of motion?
Answer: (D)
Solution
v = dx/dt = 8t³+5. At t=0, v=5 ms⁻¹; at t=1, v=13 ms⁻¹. Increase in KE = ½(2)(13²−5²) = (169−25) = 144 J.