Question

a body moving with uniform acceleration covers 12 m in 2nd second and 20m in 4th second of its motion.how much distance will it cover in 4 seconds after the 5th second​

Answer 1

Answer:

The distance will it cover in 4 seconds after the 5th second​ is 136 m

Explanation:

The body moving with uniform acceleration covers 12 m in 2nd second and 20m in 4th second of its motion.

Now, Let the initial velocity as v and the acceleration as a

so from the problem,

12=(2v+ 1/2 a(2)^2) - (v+ 1/2 a(1)^2)

  = v+ 3a/2

12-3a/2 =v

20 = (4v+ 1/2 a(4)^2) - (3v+ 1/2 a(3)^2)

      = v+ 7a/2

20 -7a/2= v

Therefore,

20 -7a/2=12-3a/2

8= 2a

a= 4

hence putting the value of a  we get

v= 3/2 x4

   = 6 m/s

Let w be the velocity of the fifth second

w = v+5a

    = 6+ 5x4

      = 26 m/s

distance will it cover in 4 seconds after the 5th second D= 4w+ 1/2a(4)^2

                                                                                                = 4x26+1/2x4x16

                                                                                                =  136 m