Question

Body A starts from rest with an acceleration a,. After 2s,
another body B starts from rest with an acceleration an.
If they travel equal distances in the 5th second, after the
start of body A, then the ratio a: az is equal to
1)5:9
2) 5:7
3) 9:5
4)9:7
answer-5:9 how??? chapter-kinematics ​

Answer 1

Answer:

=> 5:9

Explanation:

Well the answer of this question lies in a single formula

=> Sn = u + ½ a(2n-1)

where u = initial velocity

S = distance travelled

a = acceleration

n = nth time

--------------------

Given :

For body A ,

U = 0 (as starts from rest)

S = x m (Let )

n = 5th sec

a = ?

By formula

=> x = 0 + ½ a(2×5-1)

=> 2x = a(10-1) = 9a

=> a = (2/9 )x

For body B,

U = 0 (starts from rest)

a = a•

S = x

n = (5-2) = 3

By formula

=> x = 0 + ½ a•(2×3 -1)

=> 2x = a• (5)

=> a• = x(2/5)

---------------------

So the ratios of their acceleration is

=> a:a• = x(2/9) : x(2/5)

= 1/9 : 1/5

= 5:9

----------------------

Hope this is your required answer

Proud to help you

Answer 2

Answer:

the answer is 9:5 ..

its simple.

Explanation:

As we are suppose to compare a relative term(acceleration) here. i.e it varies with respect to different objects, we need to compare them with respect to a certain quantity which is same for both of the objects.

Here the Displacement fulfills our requirements.

let us try to imagine the situation with the help of this diagram:-

A   q------2sec-------w--------3sec--------e

B   q..w--- --- --- --- --- --- ----3sec----- --e

let the initial point for objects A and B be Aq and Bq respectively.

Let us assume that they travel a distance 'qw' in 2 sec and 'we' in the next 3 sec respectively.(as the object B travels no distance in the time 2sec, for it the points 'q' and 'w' coincides)

we have to consider a  same reference phase hence we consider the 'we' phase.

Now the calculations:-

displacement of A in 2 sec =

s=ut + 1/2 at^2

s=0(2) + (1/2)a×(2)^2

s=2a.........{i}

displacement of A in next 3 sec =

s=ut +1/2 at^2

s=(0)3 + 1/2 ×a(3)^2

s= 9a/2

or s = 9a/2......(ii)

displacement of B in the 3 sec =

s= ut + 1/2 at^2

s=(0)3 + 1/2 az×3^2

s=9a/2......(iii)

now from (i), (ii) and (iii)

(iii) = (ii)-(i)                 { distance traveled by B in 3 sec is equal to distance

                                   traveled by A in 3 sec minus the distance already

                                   traveled in the 2 sec }

9az/2= 9a/2 - 2a

9az = 9a - 4a

9az = 5a

so a/az = 9/5

or a:az = 9:5

Hope I got that right.