A particle travels 10m in first 5 sec and 10m in next 3 sec. Assuming constant acceleration what is the distance travelled in next 2 sec.
Dear student
Let us assume initial velocity is u and constant acceleration a.
using ,
s = ut + 1/2 a t^2
For first 5 second,
10 = 5 u + 1/2 × a × 25..........(1)
for next 3 seconds t = 8s and s = 20m
20 = 8 u + 1/2 × a × 64...........(2)
Solving 1 and 2
a = 1/3 and u = 7/6
Now for next 2 seconds t= 10s and s'
S' = 10 × 7/6 + 1/2 × (1/3) 100
S'=170/6
Now distance travelled in 2 sec = (170/6) - 20
= 50/6 => 8.3 m
Regards