Figure depicts the motion of a particle moving along an $x$ axis with a constant acceleration. What are the magnitude and direction of the particle’s acceleration ?

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$4 m / s^{2}$ in the -ve $x$ direction

$2 m / s^{2}$ in the -ve $x$ direction

$2 m / s^{2}$ in the +ve $x$ direction

$4 m / s^{2}$ in the +ve $x$ direction

answer is A.

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Detailed Solution

From the figure, we see that x₀ = –2.0 m, we can apply

$x - x_{0} = v_{0} t + \frac{1}{2} a t^{2}$

with t = 1.0 s, and then again with t = 2.0 s. This yields two equations for the two unknowns, v₀ and a:

$0.0 - (- 2.0 m) = v_{0} (1.0 s) + \frac{1}{2} a {(1.0 s)}^{2}$

$6.0 m - (- 2.0 m) = v_{0} (2.0 s) + \frac{1}{2} a {(2.0 s)}^{2}$

Solving these simultaneous equations yields the results $v_{0} = 0$ and $a = 4.0 m / s^{2}$ .

The fact that the answer is positive tells us that the acceleration vector points in the +x direction.

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