Question

A particle moving with constant acceleration of 2m/s² due west has an initial velocity of 9m/s due east.find the distance covered in fifth second of its motion??​

Answer 1

Answer:

A particle has an initial velocity of 9m/s due east and a constant accn of. A particle has an initial velocity of 9m/s due east and a constant accn of 2m/s^2 due west. The distance covered by the particle in the 5th second of its motion is .... Answer is 0.5m but i 'm getting 0.25m .

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Answer 2

Motion of Equation:

V = u + at

0 = 9 -2t

t = 4.5 sec

Now , find out velocity at 4 sec

v = u + at

v = 9 - 2 × 4 = 1m/s

so, distance covered in 4sec to 4.5 sec

= ut + 1/2at²

= 1 × 1/2 - 1/2 × 2 × 1/4

= 0.25 m

Distance = 1/2at²

=1/2 × 2 × 1/4

= 0.25 m

Distance covered in 5th second = distance covered in 4 sec to 4.5 sec + distance covered in 4.5 sec to 5sec

= 0.25m + 0.25m

= 0.5 m

Hope it will be helpful :)