Q-9- If a body starts from rest- moves with constant acceleration and travels 121 cm in nbsp-6th second- then its acceleration is nbsp-1- 0-11 m-s2-2- 2-2 m-s2-3- 0-22 m-s2-4- 1-1 m-s2

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Standard VIII

Physics

Acceleration

Q.9. If a bod...

Question

Q.9. If a body starts from rest, moves with constant acceleration and travels 121 cm in $6^{t h}$ second, then its acceleration is
(1) 0.11 $m / s^{2}$
(2) 2.2 $m / s^{2}$
(3) 0.22 $m / s^{2}$
(4) 1.1 $m / s^{2}$

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Solution

Dear Student ,
Here initial velocity= u= 0 m/s

Distance travelled = s= 121 cm= 1.21 m

Time = t= 6 sec

To find acceleration, we can use newtons 2 nd law of motion

S_t= u + 1/2 a(2t-1)

or , 1.21 = 0*6 + 11a/2

or, 1.21 = 11 a/2

or, a= 2.42/11= 0.22 m/s²

Regards

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Q.9. If a body starts from rest, moves with constant acceleration and travels 121 cm in 6th second, then its acceleration is (1) 0.11 m/s2 (2) 2.2 m/s2 (3) 0.22 m/s2 (4) 1.1 m/s2

Dear Student , Here initial velocity= u= 0 m/s Distance travelled = s= 121 cm= 1.21 m Time = t= 6 sec To find acceleration, we can use newtons 2 nd law of motion St= u + 1/2 a(2t-1) or , 1.21 = 0*6 + 11a/2 or, 1.21 = 11 a/2 or, a= 2.42/11= 0.22 m/s2 Regards

Q.9. If a body starts from rest, moves with constant acceleration and travels 121 cm in $6^{t h}$ second, then its acceleration is
(1) 0.11 $m / s^{2}$
(2) 2.2 $m / s^{2}$
(3) 0.22 $m / s^{2}$
(4) 1.1 $m / s^{2}$

Dear Student ,
Here initial velocity= u= 0 m/s

Distance travelled = s= 121 cm= 1.21 m

Time = t= 6 sec

To find acceleration, we can use newtons 2 nd law of motion

S_t= u + 1/2 a(2t-1)

or , 1.21 = 0*6 + 11a/2

or, 1.21 = 11 a/2

or, a= 2.42/11= 0.22 m/s²

Regards