Question

Particle moving under influence of a constant force is given by V =√|4-2X| where X is magnitude of displacement of the particle At t=0 initially the particle is noticed to be moving towards east.The distance travelled by the particle in first 5seconds is?
The ratio of magnitude of displacement in 4th and 5th second is?

Answer 1

so if sqrt(abs(4-2x) is displacement, you will manually change signs at x=2 of the integrand when integrating from fourth to fifth second.
a) distance= int(v dt) 0,2 + (-int(vdt)2,5

and do similar for t=4, and t=5, and remember the signs when you subtract distance@4 from distance@5