Question

A body moving with uniform acceleration describe 20 m in 2nd sec 30m in 4th sec then find distance cover by the body in 6th sec......
Here i tried to solve it with below manner......
Body moving with uniform acceleration,
S= ut+1/2at^2
For 1st condition, 20= 2u+(1/2×a×4)
2u+2a = 20-------(1)
For 2nd condition, 30= 4u+(1/2×a×16)
4u+8a = 30--------(2)
By solving (1)&(2),
a = -2.5 m/s^2
u = 12.5 m/s
I applied it for both conditions,
For t= 2sec,. S= (12.5×2) + (1/2×-2.5×4)
S = 20m
For t= 4sec, S= (12.5×4) + (1/2×-2.5×16)
S= 30m
Both given conditions are perfectly satisfied....... So I thought I was right...
But when I tried to apply it for t= 6sec
S= (12.5×6) + (1/2× -2.5 ×36)
I got S = 30m....May it be wrong.......
But why did I get the same ans of t=4sec for t= 6 sec?

Answer 1

Answer:

Given that the motion is in 1 Dimension and the body doesn't change it's direction (i.e. velocity is in the direction of acceleration), we have:

s=u+a(2t−1)/2

12 = u + a(2*2 - 1)/2

=> 12 = u + 3a/2

=> 2u + 3a = 24

and

20 = u + a(2*4 - 1)/2

20 = u + 7a/2

=> 2u + 7a = 40

From 1 and 2:

a = 4

u = 6

Now, distance covered in 4s after 5th second = Distance covered in 9 seconds - distance covered in 5 seconds

= 6*9 + 1/2 * 4 * 9 * 9 - (6 * 5 + 1/2 * 4 * 5 * 5)

= 54 + 162 - (30 + 50)

= 136 m