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
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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
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
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