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The initial velocity of a particle is 10m/s and its retardation is 2m/s^2. The distance moved by the particle in 5th second of its motion is?? a. 75m b. 50m c. 19m d. 1m

Aayush Arora , 7 Years ago
Grade 9
anser 3 Answers
Rohan

Last Activity: 7 Years ago

Sn=u+a/2 (2n-1)Sn =10-2/2 (2×5-1)10-9=1mAns d is correct please approve ans if you are comfortable

Vikas TU

Last Activity: 7 Years ago

Dear Student,
Give u a chance to be the underlying speed  as,
Furthermore, the underlying speed u = 10 m/sec 
At t = 4s 
From the newtons first eqn 
V = u +at 
As it get impeded then a = - 2m/s^2 
V4 = 10 - 8 = 2 m/sec 
At t = 5sec 
Again utilize newtons first eqn 
V5 = 10 – 2* 5 = 0 m/sec 
Normal speed in fifth second 
V = (V4 – V5)/2 = 1 m/sec 
What's more, as we realize that 
Separation = speed * time 
= 1m/sec * 1 sec = 1m.
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

Tanmay Ananf

Last Activity: 6 Years ago

Let,
t0------------------------------t4----------t5
Here,u=10m/sec
and a=-2m/sec2
First, we will calculate speed till time =4sec (0 to 4)
Therefore, t=4
since, v1= u+at
=10+(-2)(4)
=10-8=2m/sec
When t=5.
so, v2=u+at
=10+(-2)(5)
=10-10
=0m/sec
Therefore,
we can find ave. speed=v1+v2÷2
=2+0÷2
=1m/sec
So, from 4th second to 5th second there is difference of only 1 second    and there is given speed=1m/sec.
Therefore, in 1sec particle goes 1metre
SO THE ANSWER IS 1metre.
I HOPE YOU UNDERSTAND BY THIS EXPANDED FORM.
                       
              
https://www.askiitians.com/forums/General-Physics/the-initial-velocity-of-a-particle-is-10m-s-and-it_179077.htm

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