I would do it this way
find acceleration, and initial velocity.
distance covered in the fifth second is distance in the first five seconds minus distance in the first four...
d(5th)=vi(5-4)+a/2 (25-16)=vi+4.5a
76=vi+4.5a
now, the 10th second
116=vi(10-9)+a/2(100-81) or
116=vi+9.5a
so you have two equations, two unknowsn, solve for vi, and a.
Now, for the distance in the 16thru 21 seconds
d=vi(20-15)+a/2 (20^2-15^2)
A body with uniform acceleration covers 76 m in the 5th sec and 116 m in the 10th sec.Find the distance traveled in the 15th sec.what is the distance traveled in next 5 seconds.
As the body is travelling with uniform acceleration so the distance traveled by it would in 15th sec would be 156 m but how to calculate the distance in next 5 seconds
2 answers
Ghgh