Question

a particle covers 10 metre in first 5 second and 10 metre in next 3 second acceleration find initial speed acceleration and distance covered in next 2 seconds​

Answer 1

Answer:

Explanation:lets assume initial velocity is U and constant acceleration A.

using   s=ut+5 a t^2

for first 5 second

10=5u + .5x a x 25..........[1]

for next 3 seconds t= 8 and s=20

20=8u +.5xax 64.............[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 second = [170/6]-20=50/6 meter

Answer 2

Answer:

(i)Let the initial velocity = u

Total distance= 20m

Time taken= 8sec

S= ut+½ at²

20=8u+32a

20+8a=5..................(1)

(ii)For First 5seconds S=10m;t= 5sec

S= ut+½ at²

10=u (5)+½a(5)²

10=5u+25/2a

20=10u+25a

2u+5a=4....................(2)

Solving equations (1) and (2)