Question

A particle is moving in a straight line with initial velocity and uniform acceleration a. If the sum of the distance travelled in $t^{th}\ and\ (t + 1)^{th}$ seconds is 100 cm, then its velocity after t seconds, in cm/s, is(a) 80 (b) 50(c) 20 (d) 30

Answer 1

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

That's all you need, believe me.

It's the expression that gives the distance travelled by a body in the nth second, if the acceleration is uniform.

The velocity of such an object after t seconds is given by

v = u + at

So the sum of the distances travelled during the t-th and (t+1)th second amount to 100cm

Then:

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

Answer 2

Answer:

B) 50

Explanation:

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.