Question

A body starting from rest, was observed to cover 20m in 1 second and 40m during the next second. how far had it travelled before the first observation was taken?

Answer 1

Let x be the distance travelled before first observation is made.
u=0 m/s
V= ?
acceleration =a m/s²
By third equation of motion:
V²-u²=2as
V²-0=2ax
x=V²/2a-----(1)
Distance travelled when t= 1:
S=ut+ 1/2at²
20=Vx1+1/2a(1)²
40=2V+a------(2)
V1=V+at
=V+ax1
V1=V+a-----(3)
Distance travelled in next second:
S=ut+ 1/2at²
40=V1x1+1/2a(1)²
80=2V+2a+a
80= 2V+3a ---(equ 4)
By solving equation 2 and 4
40=2V+a
80= 2V+3a
- - -
-----------------
-40=-2a
a=20m/s²
2v =40-a
2v=40-20
2v=20
v=10m/s
now,X=V²/2a
=10*10/2*20
=10/4
=2.5m
∴ Distance travelled before the first observation was taken is 2.5m