A particle moves with a uniform acceleration along a straight line. It covers 41 cm and 49 cm in the 6th and the 10th seconds respectively. What will be the distance covered by the particle in 15 s?

Explanation

We know, distance covered in the nth second,
`s_(n) = u+(1)/(2)a(2n-1)`
According to the question,
`s_(6) = u+(1)/(2)a(2xx6-1) " ""or," 41 = u+(11)/(2)a" "cdots(1)`
and `s_(10)=u+(1)/(2)a(2xx10-1)" ""or,"49 = u+(19)/(2)a" "cdots(2)`
By solving equations (1) and (2) we get,
u = 30 and a = 2
Now putting t = 15 in s = `ut+(1)/(2)at^(2)` we get,
`s = 30 xx15+(1)/(2)xx2xx(15)^(2) = 450+225 = 675` cm
`:.` The particle traverses 675 cm in 15 s.