A force of 30N halves a body's velocity in 9.0m. If the body's mass is 5kg, the original velocity & time is?

Answer 1

The energy taken to slow the object from v1 to v2 is

E=FS=(1/2)m(sqr(v2)-sqr(v1))

and given v2=(1/2)(v1)

We get v1=sqrt((8/3)(FS/m)) = 12m/s

Also from v2=v1+at and F=ma

t=v1/(2a)=mv1/(2F) = 1 second

Answer 2

Lets find the deceleration factor with F=ma.
a = F / m = 30 / 5 = 6 m/s-2 (we should write MINUS 6m/s-2, since it is a DECELERATION).
"Half the velocity is attained in 9m"
Using e = a t² / 2, we can find the time it took:
t = sqr(2e / a) = sqr (2 * 9 / 6) = sqr (3).
Since v = a t, and writing:
v0 = final velocity of 0 m/s = a . t1
v1 = half velocity of vx = a . t2 = 6 sqr(3)
vx = full velocity = a . t3
we can say that:
t3 - t2 = t2 - t1 (t is directly proportional to v)
so the total time to come to a stop will be 2 x t2
giving
vx = a * 2 * t2 = 6 * 2 * sqr(3) -> initial velocity. (20.78 m/s)
the original time was 2 x t2 or 2 x sqr(3) or 3.46 s before total stop and 1.4142s before half speed.