Question

A particle starting from rest moves along a straight
line with constant acceleration. If it takes 5.0 s to
reach the speed of 18.0 km/h find (a) the average
velocity during this period, and (b) the distance
travelled by the particle during this period.​

Answer 1

Initial velocity=u=0 m/s

time=t=5sec

Final veloity=v=18km/h=18x5/18=5m/s

a) Find the average velocity during this period:

From first equation of motion :v=u+at

5=0+ax5

a=5/0.5=5/5=1m/s²

Distance =S=ut+1/at²

Average velocity=s/t=u+1/2at

Average velocity=0+1/2 x1x5

=2.5m/s

b) the distance traveled by the particle during this period  :

Distance travelled=s=Average velocity x time taken

S=2.5x5=12.5m

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Answer 2

Answer:

Explanation:

Here u=0

         t=5 seconds

         velocity=18 km/hr=18x(5/18) m/s

     so,a=v/t= 18x(5/18) /5=1 m/s

         Distance=s=1/2x a x t^2

                           =25x0.5=12.5m

Average velocity=Total displacement/Total time

                           =12.5/5=2.5 m/s