A uniformly accelarating object covers a distance of #11# meters in #6.5# seconds. If it reaches a final velocity of #3.3# m/sec, what is its initial velocity?

1 Answer
Jan 21, 2017

Answer:

Initial velocity is #0.0846# m/sec.

Explanation:

Some of the formulas used in uniformly accelerating motion with an acceleration #a#, initial velocity #u# - not known here, final velocity #v# - given here as #3.3# m/sec, time taken #t# is #6.5# sec and distance covered (in #t# sec) #S#, which is #11.0# meters, are

(A) #v=u+at#, (B) #S=ut+1/2at^2# and (C) #v^2-u^2=2aS#

From (A) we can also have #v-u=at# and dividing (C) by this,

we get #v+u=(2S)/t#

As #S=11.0# meters, #v=3.3# m/sec and #t=6.5# sec.

#3.3+u=(2xx11)/6.5=22/6.5=44/13=3.3846#

Hence #u=3.3846-3.3=0.0846#

i.e. initial velocity is #0.0846# m/sec.