Question

a particle covers 10m in first 5s and 10m in next 3s . Assuming constant acceleration. Find intial speed , acceleration and distance covered in next 2s​

Answer 1

Answer:

This is the answer

Explanation:

Let assume initial velocity is u and constant acceleration a.

Let assume initial velocity is u and constant acceleration a.using s=ut+.5 a t^2

Let assume initial velocity is u and constant acceleration a.using s=ut+.5 a t^2For first 5 second

Let assume initial velocity is u and constant acceleration a.using s=ut+.5 a t^2For first 5 second10=5 u +.5x a x 25..........(1)

Let assume initial velocity is u and constant acceleration a.using s=ut+.5 a t^2For first 5 second10=5 u +.5x a x 25..........(1)for next 3 seconds t=8 and s=20

Let assume initial velocity is u and constant acceleration a.using s=ut+.5 a t^2For first 5 second10=5 u +.5x a x 25..........(1)for next 3 seconds t=8 and s=2020=8 u+.5xax64...........(2)

Let assume initial velocity is u and constant acceleration a.using s=ut+.5 a t^2For first 5 second10=5 u +.5x a x 25..........(1)for next 3 seconds t=8 and s=2020=8 u+.5xax64...........(2)Solving 1 and 2

Let assume initial velocity is u and constant acceleration a.using s=ut+.5 a t^2For first 5 second10=5 u +.5x a x 25..........(1)for next 3 seconds t=8 and s=2020=8 u+.5xax64...........(2)Solving 1 and 2a=1/3 and u=7/6

Let assume initial velocity is u and constant acceleration a.using s=ut+.5 a t^2For first 5 second10=5 u +.5x a x 25..........(1)for next 3 seconds t=8 and s=2020=8 u+.5xax64...........(2)Solving 1 and 2a=1/3 and u=7/6Now for next 2 seconds t=10 and s'

Let assume initial velocity is u and constant acceleration a.using s=ut+.5 a t^2For first 5 second10=5 u +.5x a x 25..........(1)for next 3 seconds t=8 and s=2020=8 u+.5xax64...........(2)Solving 1 and 2a=1/3 and u=7/6Now for next 2 seconds t=10 and s'S'=10x7/6 +.5x(1/3) 100

Let assume initial velocity is u and constant acceleration a.using s=ut+.5 a t^2For first 5 second10=5 u +.5x a x 25..........(1)for next 3 seconds t=8 and s=2020=8 u+.5xax64...........(2)Solving 1 and 2a=1/3 and u=7/6Now for next 2 seconds t=10 and s'S'=10x7/6 +.5x(1/3) 100S'=170/6

Let assume initial velocity is u and constant acceleration a.using s=ut+.5 a t^2For first 5 second10=5 u +.5x a x 25..........(1)for next 3 seconds t=8 and s=2020=8 u+.5xax64...........(2)Solving 1 and 2a=1/3 and u=7/6Now for next 2 seconds t=10 and s'S'=10x7/6 +.5x(1/3) 100S'=170/6Now distance travelled in 2 sec=(170/6)-20=50/6 meter