Question

6. A particle is moving along straight line with initial velocity 48 m/sec and acceleration –10m/s2

. The

distance travelled by particle in 5th second is :
remember in 5s not after
(A) 3m (B) 115 m (C) 5

17

m (D) 5

14

m

Answer 1

your solution for your question has been provided

Answer 2

Hey user !!!

Given::-
initial velocity= 48m/s .

Acceleration= 10m/s square.


Distance=??

Theirfore ,, from 1st equation of motion,,

V= U+AT
= 0= 48+10×time
= time = 48÷10= 4.8 secs..( approx)..

NOW,, speed= distance ÷ time
= distance = speed ×time
= distance = 48×4.8
= distance= 230.4 meter..

Theirfore,, The requried ,, distance = 230.4 or 230 meter..

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