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 $x_s$ = 6.0 m. What is the particle's acceleration?

• A. - 4 m/$s^2$
• B. +4 m/$s^2$
• C. + 1 m/$s^2$
• D. + 2 m/$s^2$

Answer 1

The correct option is B

+4 m/$s^2$

The given graph is a position time graph. This means that at t = 0 the particle is at x = -2

Don't assume the displacement = -2m

That would be wrong.

Since the particle is moving with constant acceleration. That means we can use the equations of motion

Now which one to use

s = ut + $\frac{1}{2}a{t}^2$ $\because$ we have s— t relation.

Now at t =1 particle has reached x = 0

So the displacement will be 2 m

From x= — 2 to x = 0. Displacement = final position— starting position.

$\Rightarrow$ 2 = u(1) + $\frac{1}{2}a(1{)}^2$

2u + a = 4.............(1)
At t = 2 particle has reached x = 6. So after 2 secfrom start the particle is 8 cm for from starting point.

$8=u\left(2\right)+\frac{1}{2}a(2{)}^2$

$\Rightarrow$ 2u + 2a = 8

u + a = 4 .... (2)

from equation (1) and (2) we get u + 4 = 4

$\Rightarrow$ u = 0 . ... (3)

Substituting (3) in (2) get a = +4 m/$s^2$

Acceleration =4 m/$s^2$