Question

Initial velocity of particle is 10m/sec.and its retardation is 2m/sec sqaure.the distance moved by the particle in 5th sec of its motion is

Answer 1

hey here is your answer

this question can be solved by forth equation of motion

Sn = u+a/2(2n-1)
here Sn = distance covered in nth second
u = initial velocity
a = acceleration
& n = nth second

so, S5 = 10+(-2)/2(2(5)-1)
S5 = 10-1(10-1)
S5 = 10-9
S5 = 1 m

so, the particle will cover 1 meter of distance in 5th secpond of its journey

hope it helps
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