A particle covers 10m in first 5 sec and 10m in next 3 sec. Assuming constant acceleration, find initial speed, acceleration and distance covered in next 2 sec.

Explanation

Distance covered in first 5 sec = 10 m
`thereforet_(1)=5sec,s_(1)=10m`
We know `s=ut+1/2at^(2)rArr10=5u+25/2arArr10=5(u+5/2a)rArr2=u+5/2a" ".....(1)`
Distance covered in next 3 sec = 10m
`t_(2)=5+3=8sec,s=10+10=20m`
`therefores=ut+1/2at^(2)rArr20=8u+32arArr5/2=u+4a" "......(2)`
Solving (1) and (2) for a,
`{:(2=u+(5)/2a),(ul(5/2=u+4a)),(2-(5)/2=(5)/2a-4a),(-(1)/(2)=-(3)/(2)a):}`
Substitute a = `1/3m//sec^(2)` in (2)
`5/2=u+4xx1/3`
`u=5/2-4/3`
`u=7/6m//sec`
`therefore` Initial speed `u=7/6m//sec`
`a=1/2xx2/3`
`thereforea=1/3m//sec^(2)`
`therefore` Acceleration `a=1/3m//sec^(2)`
To find the distance covered in next 2 sec, we have to find the initial speed after 8 sec i.e., the final velocity after 8 sec.
`v=u+at=7/6+1/3xx8=(7+16)/6`
`thereforev=23/6m//sec`
This is the initial speed to find the distance covered in next 2 sec.
`s=ut+1/2at^(2)`
`s=7/6xx2+1/2xx1/3xx4=7/3+2/3=9/3=3m`.
`therefore` Distance covered in next 2 sec = 3m