A wheel of radius 0.160 m, which is moving initially at 47 m/s, rolls to a stop in 225 m. Calculate its linear acceleration.

To calculate the linear acceleration of the wheel, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance traveled. The equation is:

$v_f^2 = v_i^2 + 2a d$

where:

• $v_f$ is the final velocity (0 m/s, since the wheel comes to a stop),
• $v_i$ is the initial velocity (47 m/s),
• $a$ is the linear acceleration,
• $d$ is the distance traveled (225 m).

First, we rearrange the equation to solve for acceleration $a$:

$0 = v_i^2 + 2a d$

$0 = (47 \, \text{m/s})^2 + 2a (225 \, \text{m})$

$0 = 2209 \, \text{m}^2/\text{s}^2 + 450a \, \text{m}$

Next, isolate $a$:

$-2209 \, \text{m}^2/\text{s}^2 = 450a \, \text{m}$

$a = \frac{-2209 \, \text{m}^2/\text{s}^2}{450 \, \text{m}}$

$a = -4.91 \, \text{m/s}^2$

So, the linear acceleration of the wheel is $-4.91 \, \text{m/s}^2$. The negative sign indicates that the acceleration is in the direction opposite to the initial velocity, which makes sense since the wheel is decelerating.