To calculate the linear acceleration of the wheel, we can use the kinematic equation that relates initial velocity, final velocity, distance, and acceleration. 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 (225 m).

First, we rearrange the equation to solve for $ 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, meaning it is a deceleration.

Question

To calculate the linear acceleration of the wheel, we can use the kinematic equation that relates initial velocity, final velocity, distance, and acceleration. 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 (225 m).

First, we rearrange the equation to solve for $ 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, meaning it is a deceleration.

Knowee AI · Accepted Answer

To calculate the linear acceleration of the wheel, we can use the kinematic equation that relates initial velocity, final velocity, distance, and acceleration. 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 (225 m).

First, we rearrange the equation to solve for $ 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, meaning it is a deceleration.

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.

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