Question

an object return to starting point after 10 seconds if the rate of change of velocity during motion is fixed location of object after the 7 seconds will be same as that after a. 1s b. 2s c. 3s d. 4s​

Answer 1

Given :  an object return to starting point after 10 seconds if the rate of change of velocity during motion is fixed

To Find : location of object after the 7 seconds will be same as that after

a. 1s

b. 2s

c. 3s

d. 4s​

Explanation:

V = U + at

=> V = U  + 10a

V² - U² = 2aS

S = 0

=> (U  + 10a)² - U² = 0

=> U² + 100a²  + 20Ua  - U² = 0

=> 100a²  + 20Ua = 0

=>  5a + U = 0

=> a  = - U/5

S = ut + (1/2)at²

=> S = U(7)  + (1/2)(-U/5)7²

=> S = 7U  - 49U/10

=> S = 21U/10

21U/10  = Ut  + (1/2)(-U/5)t²

=> 21U  = 10Ut   - Ut²

=> 21   = 10 t   -  t²

=> t² - 10t + 21  = 0

=> (t - 7)(t - 3) = 0

=> t = 7 , t = 3

Hence at 3 s  location of object is same as after 7 sec

option C is correct

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Answer 2

HELLO DEAR,

GIVEN:- An object return to intial position at time 10 second

If the rate of change of velocity during motion is fixed,

To find the time same as position of object same as the pisition

of object at time 7 second.

SOLUTION:- By using the equation of motion

i) v = u + at

ii) v²= u² + 2as

iii) s = ut + 1/2 at²

So, v = u + at

v = u + 10a...........(1)

And , v²= u² + 2as

by putting the value of v from equation (1)

(u + 10a)² = u² + 2as

⇒ u² + 100a² + 20au = u² + 2a( 0) [ s = 0 ,because intial and final position concoid]

⇒ u² + 100a² + 20au - u² = 0

⇒ 100a² + 20au = 0

⇒ 20a ( 5a + u) = 0

⇒ a = 0 and a = (-u/5)

We take the a= (-u/5) , because for motion acceleration should be some value.

s = ut + 1/2 at²

s= u7 + ( 1/2)(-u/5) × 7² [ taking t = 7]

s= 7u - (49u / 10)

s= 21u/10

Again by using s = ut + 1/2 at²

21u/10 = ut +( 1/2 )(-u/5)t²

⇒ 21u = 10ut - ut²

⇒ 21 = 10t - t²

⇒ t² - 10t + 21 =0

⇒ t² - 7t - 3t + 21 = 0

⇒t(t - 7) -3 ( t - 7) = 0

⇒ (t-3)(t- 7) = 0

⇒ t= 3 , t= 7

Therefore , correct option is (C) 3 s.

I HOPE IT'S HELP YOU DEAR,

THANKS.