Abody of mass 40kg is accelerating at a
rate of 9m/s² . what is the rate of change of linear momentum?

Momentum = mass * velocity :  p = mv
Time Rate of change of linear momentum of an object    Δp / Δt
          = Δ (m v ) / Δt  = m Δv / Δt  = m a
         =  Force acting on the object
         =  40 Kg * 9 m /sec² = 360 Newtons

Δ denotes change in a quantity.  Δp = p2 - p1           Δt =  t2 - t1
p2  =  momentum at time t2  and          p1 = momentum at time t1
v2 = velocity at time t2              v1 = velocity at time t1
a = acceleration, constant and  same at t1 and t2
let u = velocity at time t = 0

Time rate of change of moment um is actually =  (p2 - p1) / (t2 - t1)
p2 = m v2            p1 = m v1        So  p2 - p1 = Δp  = m (v2 - v1)

v2 = u + a t2          v1 = u + a t1        so,  v2 - v1 = Δv = a (t2 - t1 )    = a  Δt

So  p2 - p1 = m (v2 - v1 )  = m a ( t2 - t1)
So            ( p2 - p1 )  / (t2 - t1 )  =  Δp / Δt  = m a  = 40 * 9 = 360 Newtons

The answer is b hope this