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A particle has an initial velocity of 9m/s due east and a constant accelaration of 2m/sec ^2 due west. the distance covered by the particle in the 5th second of its motion is? a) 0 b)0.5 c)2 d) none of the above HINT- answer is 0.5 but i keep getting 0. help me out please.

A particle has an initial velocity of 9m/s due east and a constant accelaration  of 2m/sec ^2 due west. the distance covered by the particle in the 5th second of its motion is?


a) 0


b)0.5


c)2


d) none of the above


HINT- answer is 0.5 but i keep getting 0. help me out please.

Grade:12

4 Answers

Swapnil Saxena
102 Points
12 years ago

The displacement is indeed 0 but not the distance. the distance is 0.5

Actually the the car is changing the direction of motion at 4.5 seconds. As such b/w 4 and 5 second the displacement is 0 but still the car still covers some distance.

The distance the mod of displacement made by the car b/w 4 and 4.5 seconds + mod of the displacement b/w 4.5 and 5 second. 

 b/w 4 to 4.5 seconds => (9)(0.5)-(4.52-42)= 4.5-4.25= 0.25

b/w 4.5 to 4 seconds =>  (9)(0.5)-(52-4.52)= 4.5-4.75= -0.25

Displacement =0, Distance =0.5

ABHINAV SHARMA
13 Points
5 years ago

Here the particle velocity will become zero in 4.5 seconds, as

V=U+at

0=9–2*t

t=4.5sec.

Displacement in the same period will be

v^2-U^2=2*a*S

-81=-2*S

S4.5=20.25m

Let us call the point of velocity being zero as Point A.

now body will start moving backwards,

displacement in next 0.5 sec from point A,

s=U*t+1/2*a*t^2

S0.5=1/2*(-2)*0.25= -0.25m

hence distance covered is 0.25 m

(-ve sign indicates body moved in backward direction. It is to be noted that displacement and distance are not to be confused to be same)

Distance traveled in 4 seconds

S4= 9*4+1/2*(-2)*4^2

S4=36–16=20 m

Thus,

distance covered in 5th second is

D= (S4.5+|S0.5|)-S4

D=20.25+0.25–20

D=0.5m

ankit singh
askIITians Faculty 614 Points
3 years ago

 

Here the particle velocity will become zero in 4.5 seconds, as

v=u+at0=9−2×tt=4.5sec​

Displacement in the same period will be

v2−u2=2as−81=2ss4.5​=20.25m​

Let us call the point of velocity being zero as Point A.

now body will start moving backwards,

displacement in next 0.5 sec from point A,

s=ut+21​at2s0.5​=21​(−2)(0.25)2=−0.25m​

hence distance covered is 0.25 m

(-ve sign indicates body moved in backward direction. It is to be noted that displacement and distance are not to be confused to be same)

Distance traveled in 4 seconds

s4​=9×4+21​(−2)(4)2s4​=36−16=20​

Thus,

distance covered in 5th second is

D=(s4.5​+∣s0.5​∣−s4​)=20.5+0.25−20=0.5

Keshav
13 Points
2 years ago
We know , 
S_nth= u+a(2n-1)/2
Here, in the question , u=9m/s
And a= -2 because it is in opposite side , so it is negative considered.
Now,
While applying this formula,
S_5th=9+(-2)(2(5)-1)/2
=0
When this situation arises , we solve the question by graph 
So , we think  , if displacement is 0 in 5th second , so at any time also velocity is 0 , so we find time at which v=0 , by applying eq.n of motion  , we get at t = 4.5 , v=0  , 
We know that , 5th second means , from 4 to 5 second  , so , at v=0 , the line intersect the x-axis, so by graph 
(1/2)(1/2)(1) + (1/2)(1/2)(1)
= 1/2 m =0.5m, hope you will understand , thanku

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