QuestionA skateboarder slows from $21 \mathrm{~m/s}$ to rest at $-2.5 \mathrm{~m/s}^2$ . Find the stopping distance.

Studdy Solution

STEP 1

Assumptions1. The initial velocity of the skateboarder is $21 \mathrm{~m/s}$ . The final velocity of the skateboarder is $0 \mathrm{~m/s}$ (since he comes to rest)
3. The acceleration (which is negative since it's a deceleration) is $-.5 \mathrm{~m/s}^{}$
4. We need to find the distance covered before the skateboarder comes to rest

STEP 2

We can use the equation of motion that relates final velocity, initial velocity, acceleration, and distance $v_f^2 = v_i^2 +2ad$ where $v_f$ is the final velocity, $v_i$ is the initial velocity, $a$ is the acceleration, and $d$ is the distance.

STEP 3

We can rearrange this equation to solve for distance $d$ $d = \frac{v_f^2 - v_i^2}{2a}$

STEP 4

Now, we can plug in the given values for the final velocity, initial velocity, and acceleration $d = \frac{(0 \mathrm{~m/s})^2 - (21 \mathrm{~m/s})^2}{2 \times -2. \mathrm{~m/s}^{2}}$

STEP 5

Calculate the distance $d = \frac{-441 \mathrm{~m}^{2}/\mathrm{s}^{2}}{-5 \mathrm{~m/s}^{2}} =88.2 \mathrm{~m}$ The skateboarder goes88.2 meters before he stops.

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