Figure depicts the motion of a particle moving along an x axis with a constant acceleration. The figure’s vertical scaling is set by${\mathit{x}}_{\mathbf{s}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$. What are the (a) magnitude and(b) direction of particle’s acceleration?

The kinematic equations related the initial velocity, final velocity, acceleration, displacement, and time to each other.Use these equations to find the linear relationship between initial velocity and acceleration. It is given that, $x_s=6.0\mathrm{m}$.

Using the second equation of motion,

$x-x_o=v_{o}t+\frac{1}{2}a{t}^2$

From figure, $x=0.0\mathrm{m}\mathrm{at}\mathrm{t}=1.0\mathrm{s}$.

localid="1656149026288" $0-\left(-2\right)=v_o\times 1+\frac{1}{2}a\times 1^2$

$2=v_o+\frac{1}{2}a$(i)

Also, $x=6.0\mathrm{m}\mathrm{at}t=2.0\mathrm{s}$at

$6-\left(-2\right)=v_o\times 2+\frac{1}{2}a\times 2^2$

$4=v_o+a$ (ii)

Simultaneously solving equation (i) and (ii,

$v_o=0\mathrm{m}/\mathrm{s}$

$a=4.0\mathrm{m}/{\mathrm{s}}^2$ (iii)

Therefore, the magnitude of acceleration is $4.0\mathrm{m}/{\mathrm{s}}^2$.

The acceleration found in equation (iii) is positive. Therefore, acceleration is directed in the$+x$ direction.