Question

When a force acts on a body of mass m, its position x
various with time t asx = at4 + bt + c
Where a, b and care constants. The work done by
the force during the first second of the motion is​

Answer 1

Explanation:

firstly find acceleration by double differentiating the given x derivatives

v = 4at^3 + b

a= 12at^2

now find force

F = m × 12at^2

work done = force × displacement

now put t = 1 sec in x derivative

we have x = a + b

work = (a+b) (12ma)

hope you got it