Atime-varying horizontal force f(t) = at4 + bt2 acts for 0.500 s on a 12.25-kg object, starting attime t = 1.00 s. in the si system, a has the numerical value 4.50 and b has the numerical value8.75. what impulse does this force impart to the object?

3.82 Ns

Explanation:

Time varying horizontal Force is given as

F(t) = A t⁴ + B t²

F(t) = 4.50 t⁴ + 8.75 t²

Impulse imparted is given as

impulse is 12.8614 kg

Explanation:

Given data

F(t) = At4 + Bt2

time t = 1.00 s

A = 4.50

B = 8.75

to find out

What impulse does this force impart to the object

solution

we know impulse is the change in momentum so we can right this as that

impulse I  = F i.e

dI = F(t) dt

we integrate it with limit 1 to 1.5

I =

I = +

put the value A and B

I = (4.50(1.5)^5 / 5) - (4.50(1)^5 / 5)  + (8.75(1.5)^3 / 3) -  (8.75(1)^3 / 3)

I = (4.50(1.5)^5 / 5) - (4.50(1)^5 / 5)  + (8.75(1.5)^3 / 3) -  (8.75(1)^3 / 3)

I = 6.8343 - 0.9 + 9.8437 - 2.9166

impulse is 12.8614 kg

the answer would be a

answer: a. same unit

explanation: