Question

An object is moving with uniform acceleration it's velocity after 5s is 25m/s and after 8s is 34m/s find the distance travelled by the object in 12thsecond

Answer 1

Given :-

• Velocity of object after 5sec is 25 m/s
• Velocity after 8 sec is 34 m/s

To Find :-

• The distance travelled by the object in 12th second.

Solution :-

• Let the initial velocity of the object be u and Acceleration be a .

Using Equation of Motion:-

• 25 = u+5a . . . . . . eq (1)

• 34=u+8a . . . . . . eq (2)

Now , Subtract equation (1) from (2) :-

$\sf\: \: \: \: \: \: \: \::\implies\:34-25=8a-5a$

$\sf\green{\: \: \: \: \: \: \: \::\implies\:a=3ms^{-2}}$

Putting a=3 in equation (1):-

$\sf\green{\: \: \: \: \: \: \: \::\implies 25 = u+5a}$

$\sf\: \: \: \: \: \: \: \::\implies 25 = u+5\times 3$

$\sf\: \: \: \: \: \: \: \::\implies 25 = u+15$

$\sf \: \: \: \: \: \: \: \:\green{:\implies u = 10\: m/s}$

Distance travelled in 12th second:-

$\sf\pink{\: \: \: \: \: \: \: \:S_{nth}=u+\dfrac{a}{2}(2n-1)}$

$\sf\: \: \: \: \: \: \: \::\implies \:S_{12th}=u+\dfrac{a}{2}[2(12)-1]$

$\sf\: \: \: \: \: \: \: \::\implies\:S_{12th}=10+\dfrac{3}{2}(23)$

$\sf\: \: \: \: \: \: \: \::\implies\:S_{12th}=10+\dfrac{69}{2}$

$\sf\: \: \: \: \: \: \: \::\implies\:S_{12th}=\dfrac{89}{2}$

$\sf\: \: \: \: \: \: \: \:\pink{:\implies\:S_{12th}=44.5m}$

$\therefore\:\underline{\textsf{ The distance travelled by the object in 12th sec is \textbf{44.5\: m}}}.$

Answer 2

Given : Velocity of object after 5sec is 25m/s & Velocity after 8 sec is 34m/s.

To Find : Find the distance travelled by the object in 12th second ?

_________________________

Solution : Let the initial velocity of the object be u and Acceleration be a.

$~$

$\underline{\frak{As ~we ~know ~that~:}}$

• $\boxed{\sf\pink{v~=~u~+~at}}$★

$~$

• After 5s
• ${\sf{25~=~u~+~5a\qquad\bigg\lgroup{Eqⁿ~1}\bigg\rgroup}}$
• After 8s
• ${\sf{34~=~u~+~8a\qquad\bigg\lgroup{Eqⁿ~2}\bigg\rgroup}}$

$~$

◗Subtract equation 1 from equation 2

• ${\sf{34~-~25~=~u~+~8a~-~u~-~5a}}$
• ${\sf{9~=~3a}}$
• ${\sf{a~=~3m/s^2}}$

$~$

◗Substitute a = 3m/s² in equation 1

• ${\sf{25~=~u~+~5~×~3}}$
• ${\sf{u~=~10m/s^2}}$

$~$

◗Distance travelled in 12th second

• ${\sf{S~=~u~+~a\bigg(n~-~\dfrac{1}{2}\bigg)}}$
• ${\sf{S~=~10~+~3~×~\bigg(12~-~\dfrac{1}{2}\bigg)}}$
• ${\underline{\boxed{\frak{\purple{S~=~44.5m}}}}}$★

$~$

Hence,

$\therefore\underline{\sf{Distance ~travelled~ in ~12th ~second~ is~\bf{\underline{44.5m}}}}$