A particle moves along a straight line with uniform acceleration. If it travels through distances a and b, respectively in the `l`- th and m-th seconds then prove that the distance travelled in the n-th second is `s_(n) = (a(n-m)+b(l-n))/(l-m)`

A particle starts moving from rest with uniform acceleration. It travels a distance x in first 2 seconds and distance y in the next 2 seconds. Then:

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?

A particle is moving with a uniform acceleration along a straight line AB. Its velocity at A and B are 2 m/s and 14 m/s respectively. Then

A particle starts with a velocity u with a uniform acceleration f. In the p-th, q-th and r-th seconds, it moves through distances a,b and c respectively. Prove that a(q-r)+b(r-p)+c(p-q) = 0

The acceleration of a particle starting from rest moving in a straight line with uniform acceleration is 8m//sec^2 . The time taken by the particle to move the second metre is:

A particle moves in a straight line with constant acceleration, starting from a point on the line. It moves through distances a ,b,c and d, in 0. n, 2n and 3n s respectively. Show that, (i) d-a = 3(c-b) (ii) initial velocity = (4b-3a-c)/(2n) and (iii) acceleration = (c+a-2b)/(n^(2)).

A particle, starting from rest, is in uniform acceleration. Its displacement in (P- 1) seconds and P seconds are s_(1) and s_(2) respectively . Then what would be its displacement in the (P^(2)-P+1) -th second?

A particle moves with uniform acceleration and the distance travelled in t - th second is s_(t) = 3 + 2t . Find the initial velocity of the particle .

A particle starts from rest with a uniform acceleration and travels 57 cm in the 10th second. Find out this distance travelled in 12 s and the velocity at that instant.