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 :

(A) 3m (B) 115 m (C) 5

17

m (D) 5

14

m can anyone ans it fast plzzzz remember in 5s not after

Answer 1

distance travelled in nth second:
S1-S2= u + 1/2a(2n-1)
here u=48m/s
a=10
n=5
putting these values in above formula & get the ans

Answer 2

see i think option (A) is incomplete it should be 93 and not just 3
it is so that s=(2u+at)×t/2=ut+at²/2
so to find the distance traveled in the fifth second you need to find out the difference between the distance traveled in 4s and 5s
so here in the formula u is the initial velocity,a is the acceleration, t is the time taken and s is the distance travelled
so when t is 4 ,it is......
s1=(2u+at)×t/2 =(2*48+10*4)×4/2 =(96+40)*2= 136×2=272
when t is 5 ,it is.......
s2=(2u+at)×t/2=ut+at²/2 =48*5+10*5*5/2=48*5+5*5*5=240+125=365
so the distance traveled in the fifth second is...
s2-s1=365-272=93
HOPE IT HELPS