Question

If a particle starts at an initial speed of 17m/sec and retards at 2m/s², what is the distance covered by the particle in the 9th second of motion?

Answer 1

Answer:

316 m is correct answer according to me ☺️

Answer 2

Answer:

0.5m

Explanation:

u=17m/s

a= -2m/s^2

if we take v=0, then

t will be ,

v= u + at

0=17 + (-2)t

t = 8.5 sec.

now,

s= 1/2 at^2

s= 1/2 × 2 × (1/2)^2

= 0.5 m