Question

A body is moving in a straight line with initial velocity u and uniform acceleration a. Sum of the distances travelled in n ^ (th) and (n + 1) ^ (th) seconds is 50 m. If v is its velocity after n seconds and u=5 m/s,v/u is​

Answer 1

Answer:

Distance travelled in nth second of uniformly accelerated motion =

Sn = u+a/2(2n-1)

where u is the initial velocity and a the uniform acceleration of the body.

The sum of distances travelled in T th & (T+1) th second = 100 cm

Therefore,

ST +ST+1 =100

If the particle is moving with initial velocity U and uniform acceleration A, distance travelled by the particle in Tth second will be -

100 = s(t-th) + s(t+1th)

100 = u + a(2t - 1)/2 + u + a(2t + 2 - 1)/2

100 = 2u + a(2t - 1)/2 + a(2t + 1)/2

100 = 2u + (a/2)(2t - 1 + 2t + 1)

100 = 2u + (a/2)(4t)

100 = 2u + a(2t)

100 = 2u + 2at

100 = 2(u + at)

50 = u + at

50 = v

Therefore, the velocity after t seconds is 50cm/s.