A body moving with uniform acceleration has a velocity of 12cm /a when its x coordinate is 3cm. If its x coordinates 2s later is -5cm, what is the magnitude of its acceleration

a=change velocity/time=(vf-vi)/2


vi=12
consider position:
sf=si+at
-5=3+a*2 or a=-4cm/s^2

Answer the question

To find the magnitude of the acceleration, we can use the equation of motion:

\(v^2 = u^2 + 2a(x-x_0)\)

Where:
\(v\) is the final velocity
\(u\) is the initial velocity
\(a\) is the acceleration
\(x\) is the final position
\(x_0\) is the initial position

Given:
\(u = 12 \, \text{cm/s}\)
\(x = -5 \, \text{cm}\)
\(x_0 = 3 \, \text{cm}\)

By substituting the given values into the equation, we get:

\(12^2 = u^2 + 2a(x-x_0)\)

Simplifying further:

\(144 = 144 + 2a(-5-3)\)
\(144 = 144 - 16a\)

Now, solve for \(a\):

\(0 = -16a\)
\(\Rightarrow a = 0\)

Therefore, the magnitude of the acceleration is 0.

To find the magnitude of the acceleration, we can use the equations of motion for uniformly accelerated motion. The equation that relates velocity, initial position, acceleration, and time is:

$\Delta x = v_0t + \frac{1}{2}at^2$

Where,
- $\Delta x$ is the change in position (final position - initial position)
- $v_0$ is the initial velocity
- $a$ is the acceleration
- $t$ is the time

In this case, the body's initial velocity is given as 12 cm/s and its initial position is 3 cm. After 2 seconds, the body's position is -5 cm. Let's use this information to find the magnitude of the acceleration.

1. Calculate the change in position ($\Delta x$):
$\Delta x = (-5) - 3 = -8$ cm

2. Plug in the known values into the equation of motion:
$\Delta x = v_0t + \frac{1}{2}at^2$
$-8 = (12)(2) + \frac{1}{2}a(2)^2$

3. Simplify and solve for acceleration (a):
$-8 = 24 + 2a$
$2a = -32$
$a = -16$ cm/s² (magnitude of acceleration is 16 cm/s²)

Therefore, the magnitude of the acceleration is 16 cm/s².