Question

A particle has initial velocity of 17m/s
towards east and constant acceleration
of 2m/s due west. The distance covered by
it in 9th second of motion is .​

Answer 1

Answer:

34 m is your answer.

Explanation:

Given,

u=17m/s

a=2m/s

By using this formula,

S of nth second=u+a/2(2n-1) we can find the answer,

Now,

s of 9th second=17+2/2(2×9-1)

=17+2/2(18-1)

=17+2/2(17)

=17+17

=34 m

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