A particle travels 10 m in first 5 secs, 10 m in next 3 secs. assuming constant acceleration, what is the distance travelled in next 2 sec?

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

Answers

2016-03-09T12:08:16+05:30

Let assume initial velocity is u and constant acceleration a.

using    s=ut+.5 a t^2

For first 5 second

10=5 u +.5x a x 25..........(1)

for next 3 seconds t=8 and s=20

20=8 u+.5xax64...........(2)

Solving 1  and 2

a=1/3 and u=7/6

Now for next 2 seconds t=10 and s'

S'=10x7/6 +.5x(1/3) 100

S'=170/6

Now distance travelled in 2 sec=(170/6)-20=50/6 meter

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