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chemistry100

A car accelerates from rest at constant rate of 2 ms^-2 for sometime. Then, it retards, at constant rate of 4 ms^-2 and comes to rest. If the total time for which it remains in motion be 3 seconds, what is total distance traveled

(a) 4 m (b) 6 m (c) 2 m (d) 3 m

astatine19

hmm... i think the data is insufficient...

akhil_o

let t1 be time of acc,t2 be time of deceleration

v=a(t₁)=2t₁,

also 0-v=-r(t₂)=4t₂

hence ==>2t1=4t2 or t1=2t2

Also t1+t2=3

from these we get t1=2sec,t2=1 sec

akhil_o

s1=1/2*2*(2)²=4 m

v=2*2=4m/s

s2=4(1)-1/2*4*(1)(1)

=4-2=2m

Hence total s=4+2=6 m

fwks_phoenix

A car accelerates from rest at constant rate of 2 ms^-2 for sometime.

Then, it retards, at constant rate of 4 ms^-2 and comes to rest.

If the total time for which it remains in motion be 3 seconds, what is total distance traveled

all i could get was distance travelled in first half is d same as distance travelled in second half and total distance is [3(time in first half)^2]/2

just use the three basic formulae : v=u+at

v2=u2+2as

s=ut+1/2(at2)

in the first half, put u=0 and find out v and s in terms of t

and then v of first half will b u of second half and v of second half will b 0.

raulrag009

draw v-t graph

let v be max velocity attained by it

area under v-t curve gives distance travelled

therefore

s1=1/2 * v * t1

s2=1/2 * v* t2

add both

s=s1+s2=1/2*v(t1+t2)=v/2*3 =3v/2 .........[1]

{as t1+t2=3}

now using

equation of motion

v-0=2t1

and

0-v=-4t2

simplify

we get t1=v/2 and t2=v/4

add both t1+t2=3v/4

As t1+t2=3 therefore 3v/4=3 ,hence v=4

put v=4 in eqn [1]

we get distance = {3*4)/2=6m

rate if useful

elessar_iitkgp

This problem is best solved by a v-t graph

The slope of the first line is 'a' and that of the second is "-b"
Let the maximun velocity be v_m
Then, v_m/T = a

v_m/a = T ---------(1)
& v_m/t-T = b

v_m/b = t-T ---------(2)
Adding (1) and(2)
v_m = abt/(a+b)

As the v-t graph is entirely above the t axis,
So distance travelled = Displacement = Area under v-t curve = (1/2)v_mt
=abt² /2(a+b)

Substitute the values a = 2, b =4, and t = 3

deeplove

Here, the net change in the velocity over the journey is zero hence the area under the a-t curve i.e.

a.dt = 0   .......................................(1)

Let t be the time where the velocity is maximum (the time at which the acceleration changes sign)

by (1), we have   2t = 4(3-t)
                                    t=2    .....................................(2)

Hence v(2) = 2*2 = 4

Hence S = x1 + x2 = (1/2)2*2² + 4*1 - (1/2)*4*1²

                            =6m. ANS.

astatine19

whoops, didn't see the other conditions ( comes to rest and starts from rest) sorry.

ramyani

Let the train starts with an accln "a " from one station ( starting from rest ) and goes for some time t₁ . Then maxm vel v = a t₁ .

Let after t₁ , the brake is applied and the train retards at rate " b " for rest ( t-t₁ ) time and finally comes to rest . { t = total time }

Then 0 = v - b ( t-t₁ ) = a t₁ - b ( t-t₁ )

or t₁ = bt / ( a + b )

So, maxm vel v = abt / ( a + b )

Now draw a v vs t graph. ( as done by elessar_iitkgp sir )

Remember distance travelled = area under v-t graph

                                           = ( 1 / 2) ( maxm vel ) ( total time )

                                          = ( 1 / 2) [ abt / ( a + b ) ] t

                                         = ( 1 / 2) [ abt² / ( a + b ) ]

ramyani

of course elessar_iitkgp (1907) sir's method is more compact and better.

pannaguma

ans b.

rv_hbk

Its simple!!!

First , suppose the car acclerates for time t₁to attain a max. velocity given by

v=0+2*t₁=2t₁

For the retarding motion,

final vel=0.

Let time taken from max. velocity to 0=t₂

Hence,0=2t₁- 4t₂

ALSO,

t₁+ t₂ =3

Solving these 2,we get

t₁=2 & t₂=1

Total distance covered=0+1/2*2*(t₁)² + 2t₁- 1/2*4*(t₂)²

= 6m

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