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A particle is moving along the straight line with an initial velocity 48m/sec and acceleration -10m/sec 2 .the distance traveled by particle in 5 th second

A particle is moving along the straight line with an initial velocity 48m/sec and acceleration -10m/sec2 .the distance traveled by particle in 5th second

Grade:11

2 Answers

Gowtham
40 Points
6 years ago
formula for the distance travelled by a body in nth second is Sn=u+a/2(2n-1)
n=5
intial velocity u=48
it is retarding so a
s5=3m
 
FUN BLASTER
26 Points
2 years ago
In this question firstly we will check at which instant velocity of particle become zero 
u=48m/s
a=-10m/s²
 
v=u+at
0=48-10×t
t=4.8s
This is in between our 5th sec.
So firstly find distance till 4.8s than subtract from distance till 4sec .
 
Till 4.8seconds
v²=u²+2as
0=(48)²-10s
S=115.2m
 
 
Till 4 sec 
S=ut+1/2at²
S=48×4-1/2×10×4²
S=112
 
Distance 4.8s-4sec = 115.2-112=3.2m = distance from starting of 5th sec till 4.8s
 
Now we find distance in remaining 0.2sec from 5sec
t=0.2sec
u=0
a=+10m/s
S=48(0.2)+1/2×10×(0.2)²
S=0.2m
 
At last total distance in 5th sec = distance from starting of 5th sec till 4.8s + distance of  remaining 0.2s
 
= 3.2+0.2m=17/5m

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