The magnitude of the net force exerted in the x direction on a 2.50-kg particle varies with the time as shown in fig., find (a) the impules of the force. (b) the final velocity the particle attains if it is originally at rest,
(c ) its final velocity if its original velocity is `-2.0ms^(-1)`, and
(d) the average force exerted on the particle for the time interval between 0 and 5 s.

Explanation

(a) The impulse is to the right and equal to the area under the `F-t` graph `I={[5+1]//2} xx 4 = 12 Ns`
(b) `mvec(v)_(i)+vec(F)t = mvec(v)_(f)`
`implies 2.5 xx 0 + 12 = (2.5)v implies v=4.8 ms^(-1)`
( c) From the same equation,
`rArr 2.5 xx (-2) + 12 = (2.5)v rArr v = 4.8 m s^(-1)`
(d) F_(avg)Delta t=12.0implies F_(avg)=2.40N`.