Question

particle covers 10 metre in first 5 seconds and 10 metres in next 3 seconds asuming constant acceleration find initial speed acceleration and distance covered in next 2 seconds.​

Answer 1

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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

Answer 2

Hey mate here is your answer

Answer will be 50/6m

$solution$

let,assume initial velocity is U and constant accleration A using S=ut+(0.5)at square

For first 5 seconds

10=5u+(0.5)a(25)

for next 3 seconds, t=8 and s=20

20=8u+(0.5)a(64)

solving

a=1/3 and u=7/6

for next 2 seconds, t=10 and s power 1

s power 1=10(7/6)+(0.5)1/3(100)

s power 1=170/6

now distance travelled in 2 seconds

170/6-20=50/6 meter

Hope it helps you

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