Question

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

Answer 1

The magnitude of the acceleration of the body is 2.35 m/s^2.

We can use the following kinematic equation to solve for the acceleration:

x = x0 + v0t + 1/2 at^2

where x is the final position, x0 is the initial position, v0 is the initial velocity, t is the time, and a is the acceleration.

Plugging in the given values, we get:

-0.05 m = 0.03 m + 12 m/s * 2 s + 1/2 * a * (2 s)^2

Simplifying and solving for a, we get:

a = (0.05 m - 0.72 m)/2 s^2 = -0.335 m/s^2

Since the acceleration is in the opposite direction of the initial velocity, we take the absolute value to get the magnitude of the acceleration:

|a| = 0.335 m/s^2 ≈ 2.35 m/s^2.

Answer 2

The magnitude of acceleration can be calculated using the following kinematic equation:

x = x0 + v0t + 1/2at^2

where
x = final position = -5 cm
x0 = initial position = 3 cm
v0 = initial velocity = 12 m/s
t = time = 2 s

Converting all units to SI units, we get:

x = -0.05 m
x0 = 0.03 m
v0 = 12 m/s
t = 2 s

Substituting these values into the equation and solving for a, we get:

a = 2(x - x0 - v0t) / t^2
a = 2(-0.05 - 0.03 - 12(2)) / (2)^2
a = -12.5 m/s^2

Therefore, the magnitude of acceleration is 12.5 m/s^2.