A particle is moving in a straight line with initial velocity u and uniform acceleration a . If the sum of distances travelled in Tth and (T + 1)th second is 100 cm then its velocity after t seconds in cm/s is ?
A particle is moving in a straight line with initial velocity u and uniform acceleration a . If the sum of distances travelled in Tth and (T + 1)th second is 100 cm then its velocity after t seconds in cm/s is ?
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+1)th]
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.
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+1)th]
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.
