Answer :
The velocity at the end of the 7th second is 1.198 m/s
Explanation:
Assuming that the motion of the body is a uniformly accelerated motion (=constant acceleration), we can use the following suvat equation:
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
s is the displacement
u is the initial velocity
t is the time elapsed
a is the acceleration
In the first 2 seconds, the body travels 2 m, so we have:
[tex]s_1 = 2 m\\t_1 = 2 s[/tex]
So the equation can be written as
[tex]s_1=ut_1 + \frac{1}{2}at_1^2\\2=2u+2a[/tex]
If we then consider the whole first 4 seconds of motion, we have:
[tex]s_2 = 2 + 2.2 = 4.2 m\\t_2 = 2+2 = 4 s[/tex]
So the equation can be written as
[tex]s_2=ut_2+\frac{1}{2}at_2^2\\4.2 = 4u+8a[/tex]
So we have a system of 2 equations in 2 variables:
[tex]2=2u+2a\\4.2=4u+8a[/tex]
By multiplying the 1st equation by 2 and subtracting eq.(1) from eq.(2), we get
[tex]0.2=6a\\ \rightarrow a=\frac{0.2}{6}=0.033 m/s^2[/tex]
And susbtituting into the 1st equation, we get
[tex]u=1-a=0.967 m/s[/tex]
Now we can apply the following suvat equation:
[tex]v=u+at[/tex]
And substituting t = 7 s, we find the velocity at the end of the 7th second:
[tex]v=0.967+(0.033)(7)=1.198 m/s[/tex]
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