Question

If a runner with a certain initial velocity moves with uniformely acceleration in such a way that he covers 200 cm in the 2nd second and 220 cm in the 4th second,then find his initial velocity and acceleration

Answer 1

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

For 2nd second
200 = u + a(2 - 1/2)
200 = u + 3a/2 ——-(1)

For 4th second
220 = u + a(4 - 1/2)
220 = u + 7a/2 ——(2)

Subtract equation (1) from equation (2)
220 - 200 = u + 7a/2 - u - 3a/2
20 = 2a
a = 10 cms^2

Substitute a = 10 cms^2 in equation (1)
200 = u + 3*10/2
u = 200 - 15
u = 185cm/s

Initial velocity = 185cm/s
Acceleration = 10cm/s^2