Question

A particle has an initial velocity of 9 m/s due east
and a constant acceleration of 2m/s2 due west. The
distance covered by the particle in the fifth second
of its motion is:​

Answer 1

Answer:

18 meters

Explanation:

S= Vt + 1/2 at^2

S5= 9×5+2/2(5)^2

S5= 45 + 1(25)

S5= 70......(1)

similarly for fourth second

S4= Vt + 1/2 at^2

S4= 9×4 + 2/2(4)^2

S4= 36 + 1(16)

S4= 52.......(2)

Now S= S5-S4

S= 70-52

S= 18...... answer