IIT JEE , AIEEE Forum - Physics , Mechanics - A particle starts frm rest wid accn 2m/s2. accn of the particle decreases

zoomtoarpita

1.A particle starts frm rest wid accn 2m/s2. accn of the particle decreases down to 0 uniformly in 4s.Find the dist. travelled by the particle durin time interval of 4s

a.8m

b.20m

c.4m

d.2m

2. A pt. moves in a st. line under retardation k ${v}^{2}$ . if initial velocity is u, dist. covered in t sec is

a.kut

b. 1/k logkut

c.1/klog(1+kut)

d. klogut

plz gimme detailed soln

celestine

1. im gettin 32\3 m check ur question please

2. Its C . dv\dt = -kv^2 = v dv\dx
solve it to get ans

kbanagar

for the first one is the answer......8m.

chk by xt graph u will get the answer.........

or shud i giv the detailed xpln in graphical terms.........

celestine

k banagar how are u tellin its 8

celestine

and how did u draw the graph

akshay.khare91

1 .

is the ans. 8 m

if yes i can give u soltions

2 .

c

kbanagar

i'm sorry the ans is 4m..............

kbanagar

see tht in a xt graph slope corresponds to accn

accn {positve slope} = 2 s.i. units

this implies y/x = 2/ 1 =2

let whatever be negative retardation body shud come to rest.............

so a triangle is formed with height 2 units n base is 4

area of the triangle = 1/2 (2)(4) = 4 m

shinee

initial acceleration=2m/s^2
acceleration of the particle decreases uniformly in 4 seconds
(we can compare the analogy with uniform decrease in velocity using the formula 0=u-at )
where we can take 0 as the final acceleration, u as initial acceleration and a as the rate of change of acceleration
so, 0=2-ai4(where ai is the rate of change of acceleration)
ai=0.5
substituting in the formula s=(v^2-u^2)/(2ai), (where s is the velocity, v is the final acceleration, u , the initial acceleartion and ai, the arte of change in acceleration)
velocity=4/(2X0.5) =4m/s
since this is the final velocity is 0, and initial velocity is 4m/s we can find out the distance traveled using the formula distance=avg velocityXtime
(0+4)X4/2=8m

celestine

kbanagar slope of xt graph is VELOCITY

Shinee ur approach in analogy with velocity i only approximately correct though it gets the ans within the options

Here you can solve it usin integration

dA/dt = .5 is rate of change of acc
acc at any time is 2 - .5 t
vel at any time is 2t - .25t^2
dist travelled is t^2 - (t^3 /12)
when t is 4 dist is 32/3
I hope you agree????

RyuAmakusa

@ shinee that formula u have used is valid only for cont. acc.
celestine ur ans in correct.i got the same

shinee

ya, i agree with u, sorry zoomtoarpita, i have used a formula of uniform acceleration

kmmankad

i agree wid celestine!

sboosy

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