Question

A motorcyclist moving with uniform retardation takes 10s and 20s to travel successive quater kilometre. How much further he will travel before coming to rest? (Ans: 10.42m)

Answer 1

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Please refer to the answer given below :

Let 'u' be the initial velocity of the motorcycle when it covers first 250 m in 10 s and let 'a' be the retarding acceleration of the vehicle .

Let v be the final velocity after 10 s.

Since we know that, v= u +at   [ Kinematical equation ]

Substituting the value of t as 10 s, we get,

v = u +10a .................. [ eq. 1 ]

Also, we know that, s = ut + $\frac{1}{2}$ × a × t²

⇒ 250 = 10u + $\frac{1}{2}$ × a × (10)²

⇒ 250 = 10u + 50a

⇒ 25 = u + 5a .......................[ eq. 2 ]

After 20 s , let u' be the initial velocity after 20 s which will be equal to u + 10a.

∴ Again using the equation s = u't + $\frac{1}{2}$ × a × t²

and substituting s = 250 , u' = u + 10a , & t = 20 s, we get,

⇒ 250 = (u + 10a) × 20 + $\frac{1}{2}$ × a × ( 20 )² 

⇒ 250 = 20u + 200a + 200a

⇒ 250 = 20u + 400a

⇒ 25 = 2u + 40a ................[ eq. 3 ]

Multiplying '2' in equation 2 and subtracting eq. 2 from eq. 3, we get , 

⇒ 25 = 2u + 40a
    50 = 2u + 10a
    -      -      -
_______________
   -25 = 0 + 30a

⇒ a = -25/30

⇒ a = - 5/6 m/s²

Substituting a in eq. 2, we get,

⇒ 25 = u + (-5/6)×5

⇒ 25 + 25/6 = u

⇒ 175/6 m/s = u

Let v' be the final velocity after 20 s.

∴ v' = u' + at

Substituting u' = u + 10a and t = 20 s, we get,

⇒ v' = u + 10a + 20a

⇒ v' = u + 30a

Let ' s' ' be the distance travelled after 20 s beofre coming to rest.

∵ It finally comes to rest, therefore its final velocity is 0 .

Also, we know that, v² = u² + 2as.

Substituting v = 0 m/s , u = u + 30a , we get,

⇒ 0 = ( u + 30a )² + 2as'

Again substituting u = 175/6 m/s and a = -5/6 m/s² in the above equation, we get,

⇒ 0 = ( 175/6 - 30×5/6 )² - 2 × 5/6 × s'

⇒ 0 = ( 175/6 - 25 )² - 5/3 × s'

⇒ 0 =  ( 25/6) - 5/3 × s'

On solving further, s' = 125/12 i.e.,  10.42 m

This is your answer.

Hope this HELPS !!!

Answer 2

Let 'u' be the initial velocity of the motorcycle when it covers first 250 m in 10 s and let 'a' be the retarding acceleration of the vehicle .

Let v be the final velocity after 10 s.

Since we know that, v= u +at

Substituting the value of t as 10 s, we get,

v = u +10a .................. [ eq. 1 ]

Also, we know that, s = ut + 1/2 × a × t²

⇒ 250 = 10u + 1/2 × a × (10)²

⇒ 250 = 10u + 50a

⇒ 25 = u + 5a .......................[ eq. 2 ]

After 20 s , let u' be the initial velocity after 20 s which will be equal to u + 10a.

∴ Again using the equation s = u't + 1/2× a × t²

and substituting s = 250 , u' = u + 10a , & t = 20 s, we get,

⇒ 250 = (u + 10a) × 20 + 1/2 × a × ( 20 )²

⇒ 250 = 20u + 200a + 200a

⇒ 250 = 20u + 400a

⇒ 25 = 2u + 40a ................[ eq. 3 ]

Multiplying '2' in equation 2 and subtracting eq. 2 from eq. 3, we get ,

⇒ 25 = 2u + 40a

50 = 2u + 10a

- - -

______________

-25 = 0 + 30a

⇒ a = -25/30

⇒ a = - 5/6 m/s²

Substituting a in eq. 2, we get,

⇒ 25 = u + (-5/6)×5

⇒ 25 + 25/6 = u

⇒ 175/6 m/s = u

Let v' be the final velocity after 20 s.

∴ v' = u' + at

Substituting u' = u + 10a and t = 20 s, we get,

⇒ v' = u + 10a + 20a

⇒ v' = u + 30a

Let ' s' ' be the distance travelled after 20 s beofre coming to rest.

∵ It finally comes to rest, therefore its final velocity is 0 .

Also, we know that, v² = u² + 2as.

Substituting v = 0 m/s , u = u + 30a , we get,

⇒ 0 = ( u + 30a )² + 2as'

Again substituting u = 175/6 m/s and a = -5/6 m/s² in the above equation, we get,

⇒ 0 = ( 175/6 - 30×5/6 )² - 2 × 5/6 × s'

⇒ 0 = ( 175/6 - 25 )² - 5/3 × s'

⇒ 0 = ( 25/6) - 5/3 × s'

On solving further, s' = 125/12 i.e., 10.42 m

This is your answer.

Hope this HELPS !!!