Question

A particle starting from rest moves with constant acceleration. If it takes 5.0 s to reach the speed 18 km/h find (a) the average velocity during this period , and (b) the distance traveled by the particle during this period Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics

Answer 1

Solution :

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

Answer 2

HEY!!

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✔Initial velocity of the particle(u) = 0

✔Final velocity of the particle, v = 18 km/h = 5m/s

▶Time(t)  = 5s

▶Acceleration(a) = (v − u)/t

▶a = (5 − 0)/5 = 1 m/s^2

▶Distance (s) =ut+1/2at^2

▶s=12×1×(5×5)=12.5 m

a) Average velocity =Total displacement/Total time taken

▶Average =(12.5)/5 = 2.5 m/s

b) Distance travelled, s = 12.5 m