Question

a body leaves the origin with an initial speed of 4m / s and acceleration 6m/s2. calculates the distance traveled in one second, second and third.

Answer 1

ok now second equation of motion can be used
\[x=x _{0}+v _{0}t+\frac{1}{2}at^2\]
here
xo=0 (starts at origin
t=1 sec
vo=4m/sec
a=6m/sec^2
put in the above equation
\[x=0+4*1+\frac{1}{2}(6)(1)^2\]
this is the distance traveled in meters in 1 second
i hope u can do this

Answer 2

nd 2s
X=14m
nd 3 s
x=21m

Answer 3

do it again :) remeber it is t^2 not just t:)

Answer 4

2s
x=20
3s
x=39

Answer 5

correct you got it now :)