(2)-12 m/s(3) 42 m/s(4)9 m/lsC.7. If velociity oft a particle is given

1.	(2)-12 m/s(3) 42 m/s(4)9 m/lsC.7. If velociity oft a particle is given by V 10+2t m/s. The average acceleration between 2 a5 s is(1) 2 m/s(2) 4 m/s?(3) 12 m/s(4) 14 m/se distance cov ered by the body in time't is proportional to the square of the time t. The acceleratiof the body is(1) increasing(2) decreasing(3) zero(4) constant
Answer» c 7 v = 10+2t² => dv/dt = a = 4t now average acceleration in 2-5 sec = Integration of 4t from t = 2 ,to t = 5, and Divide by total time = 3 sec => 4t²/2 = 2t² from t = 2 to t= 5=> [2(5)²-2(2)²]/3 =(50- 8)/3 = 42/3 = 14m/s² option 4

(2)-12 m/s(3) 42 m/s(4)9 m/lsC.7. If velociity oft a particle is given by V 10+2t m/s. The average acceleration between 2 a5 s is(1) 2 m/s(2) 4 m/s?(3) 12 m/s(4) 14 m/se distance cov ered by the body in time't is proportional to the square of the time t. The acceleratiof the body is(1) increasing(2) decreasing(3) zero(4) constant

Answer»

c 7 v = 10+2t² => dv/dt = a = 4t

now average acceleration in 2-5 sec = Integration of 4t from t = 2 ,to t = 5, and Divide by total time = 3 sec

=> 4t²/2 = 2t² from t = 2 to t= 5=> [2(5)²-2(2)²]/3 =(50- 8)/3 = 42/3 = 14m/s²

option 4