Question

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?

Short answer

a) The magnitude of the acceleration is $4.0\mathrm{m}/{\mathrm{s}}^2$.

b) The acceleration vector points in the $+\mathrm{x}$direction.

Step-by-step solution

Finding the relationship between Vo and a

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)

(a) Calculating the magnitude of acceleration

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 $4.0\mathrm{m}/{\mathrm{s}}^2$.

(b) Finding the direction of acceleration

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