Answer:
3.0 m/s
Explanation:
The equation of motion for constant acceleration (a) is ...
[tex]x(t)=\dfrac{1}{2}at^2 +v_0t+x_0[/tex]
The problem statement tells us v₀ = 1, and we read from the graph that x₀=1. We also read from the graph that x(10) = 21. Filling these values into the equation, we can find a and x'(10).
[tex]21 = \dfrac{1}{2}a(10^2)+1(10)+1\\\\10=50a\qquad\text{subtract 11}\\\\a=\dfrac{1}{5}\\\\x'(t)=at+v_0\\\\x'(10)=\dfrac{1}{5}\cdot 10+1=3.0[/tex]
The final velocity of the object at t=10 s is about 3.0 m/s.
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Comment on the graph
We note this graph better represents increasing acceleration than it does constant acceleration. x(2) = 3.4 per the equation. It is graphed as about 4.
