$\begin{array}{1 1} -0.1 \;m/s^2,+5m \\ -0.2 \;m/s^2,+6m \\ 0.1 \;m/s^2,-5m \\-0.7 \;m/s^2,+5m\end{array} $

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We can use the first of our kinematic equation to solve for acceleration in terms of initial velocity, final velocity and time, all of which we know . Assuming the velocity to be positive the acceleration is :

$\large\frac{\overrightarrow{v} - \overrightarrow{v_0}}{t} $$=\overrightarrow{a} = \large\frac{0 m/s -1 m/s}{10 s}$$=-0.1 m/s^2$

From the second kinematic equation , which yields displacement fgrom initial velocity, acceleration and time , the distance traveled is:

$\overrightarrow{x}-\overrightarrow{x_0}=\overrightarrow{v}_0 t + \large\frac{1}{2}$$ \overrightarrow{a}t^2 =(1m/s) (10 s) + \large\frac{1}{2} $$(-0.1 m/s^2)(10 \;s)^2=+5 m $

Why you take velocity 1m/s inquestion it is given41m/s

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