Question

$\huge\bf {\underline {\underline{ \pink{Question : - }}}}$
For an object moving with uniform acceleration, travelling 50m in 5th sec, 70m in 7th sec.

A) Its initial velocity is 5 m/s

B) Its average velocity during 9th sec is 90 m/s

C) It travels 100 m in 9th sec

D) Both A & B

[ Hint :- Correct answer is option-(D) ]

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Answer 1

11th/Physics

Kinematics

Answer :

⧪ Distance covered by moving object in nth second of journey is given by

• Sn = u + a/2(2n - 1)

★ Distance covered in 5th second :

⭆ S(5) = u + a/2[(2×5) - 1]

⭆ 50 = u + a/2(10 - 1)

⭆ 50 = u + 9a/2 ... (I)

★ Distance covered in 7th second :

⭆ S(7) = u + a/2[(2×7) - 1]

⭆ 70 = u + a/2(14 - 1)

⭆ 70 = u + 13a/2 ... (II)

On solving (I) and (II), we get

➠ 20 = 4a/2

➠ 20 = 2a

➠ a = 10m/s² [Acceleration]

From equation (I)

➙ 50 = u + 9a/2

➙ 50 = u + (9×10)/2

➙ 50 = u + 90/2

➙ u = 50 - 45

➙ u = 5m/s [Initial velocity]

Hence, (A) is correct!

★ Ave. velocity during 9th second :

• Average velocity is defined as the ratio of total distance travelled to the total time taken. So we need to find distance travelled by body in 9th$.$

➤ S(9) = u + a/2(2n - 1)

➤ S(9) = 5 + 10/2[(2×9) - 1]

➤ S(9) = 5 + 5(17)

➤ S(9) = 90 m

Calculation of average velocity :

➳ V = S(9) / time

➳ V = 90/1

➳ V = 90m/s

Hence, (B) is also correct!

Distance covered by object in 9th second is 90m therefore (C) is incorrect.

★ Option (D) is the final answer!

Cheers!

Answer 2

EXPLANATION:-

We know that distance covered by a moving object in nth second:-

$S_(n_)=u+\frac{a}{2}(2n-1)}$

⇒Distance covered in 5th second

$S_(5)=u+\frac{a}{2}(2*5-1)}$

$S_(5)=u+\frac{a}{2}(9)}$

$50=u+\frac{9a}{2}-----EQ-1$

⇒Distance covered in 7th second:-

$S_(7)=u+\frac{a}{2}(2*7-1)}$

$S_(7)=u+\frac{a}{2}(13)}$

$70=u+\frac{13a}{2}}----\:EQ-2$

Let's SUBTRACTING EQ-1 and EQ-2

$2a=20\\\\a=\frac{20}{2}\\\\a=10$

So the acceleration is 10m/s²

Substituting it in EQ-1 we get:-

$50=u+\frac{9a}{2}\\\\50=u+\frac{9*10}{2}\\\\50=u+\frac{90}{2}\\\\50=u+45\\\\u=50-45\\\\u=5\:m/s$

So the initial velocity(u)=5 m/s

So opt A is correct

Now let's calculate the average velocity for 9th second

Distance covered by it in 9th second:-

$S_(9)=5+\frac{10}{2}(2*9-1)}\\\\S_(9)=5+\frac{10}{2}(17)\\\\S_(9)=5+5(17)\\\\S_(9)=90\:m$

So distance covered in 9th second=90 m

Now using the formula for average velocity:-

$Average\:velocity=\frac{Total\:distance}{total\:time}$

Here Total distance in 9th second =90 m

and total time = 1 sec

$Average\:velocity=\frac{90}1}$

$Average\:velocity=90\:m/s$

So Opt-B is also correct

SO opt-A and B both are correct

So the answer will be {OPT-D} Both A & B