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College Physics by Openstax Chapter 2 Problem 36

An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of 0.150 m/s² as it goes through. The station is 210 m long.

(a) How long is the nose of the train in the station?

(b) How fast is it going when the nose leaves the station?

(c) If the train is 130 m long, when does the end of the train leave the station?

(d) What is the velocity of the end of the train as it leaves?

Solution:

Part A

We are given the following: $v_0=22.0\:\text{m/s}$ ; $a=-0.150\:\text{m/s}^2$ ; and $\Delta x=210\:\text{m}$

We are required to solve for time, $t$ . We are going to use the formula

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

Substituting the given values, we have

\begin{align*} \Delta x & =v_0t+\frac{1}{2}at^2 \\ 210\:\text{m} & =\left(22.0\:\text{m/s}\right)t+\frac{1}{2}\left(-0.150\:\text{m/s}^2\right)t^2 \end{align*}

If we simplify and rearrange the terms into a general quadratic equation, we have

0.075t^2-0.22t+210=0

Solve for $t$ using the quadratic formula. We are given $a=0.075;\:b=-22t;\:c=210$ .

\begin{align*} t & =\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ t& =\frac{-\left(-22\right)\pm \sqrt{\left(-22\right)^2-4\left(0.075\right)\left(210\right)}}{2\left(0.075\right)}\\ \end{align*}

There are two values of $t$ that can satisfy the quadratic equation.

t=9.88\ \text{s} \qquad \text{and} \qquad t=283.46 \ \text{s}

Discard $t=283.46 \ \text{s}$ as it can be seen from the problem that this is not a feasible solution. So, we have

t=9.88\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

We have the same given values from part a. We are going to solve $v_f$ in the formula

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

So we have

\begin{align*} \left(v_f\right)^2 & =\left(v_0\right)^2+2a\Delta x \\ v_f & =\sqrt{\left(v_0\right)^2+2a\Delta x} \\ v_f & =\sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(210\:\text{m}\right)} \\ v_f & =20.6\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

We are given the same values as in the previous parts, except that for the value of $\Delta x$ since we should incorporate the length of the train. For the distance, $\Delta x$ , we have

\begin{align*} \Delta x & =210\:\text{m}+130\:\text{m} \\ \Delta x & = 340 \ \text{m} \end{align*}

To solve for time $t$ , we are going to use the formula

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

If we rearrange the formula into a general quadratic equation and solve for $t$ using the quadratic formula, we come up with

t=\frac{-v_0\pm \sqrt{v_0^2+2ax}}{a}

Substituting the given values:

\begin{align*} t & =\frac{-v_0\pm \sqrt{v_0^2+2ax}}{a} \\ t & =\frac{-\left(22.0\:\text{m/s}\right)\pm \sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(340\:\text{m}\right)}}{-0.150\:\text{m/s}^2} \\ t &=16.4\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part D

We have the same given values. We are going to solve for $v_f$ in the equation

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

Substituting the given values:

\begin{align*} \left(v_f\right)^2 & =\left(v_0\right)^2+2a\Delta x \\ v_f & =\sqrt{\left(v_0\right)^2+2a\Delta x} \\ v_f & =\sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(340\:\text{m}\right)} \\ v_f & =19.5\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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College Physics by Openstax Chapter 2 Problem 35

Consider a grey squirrel falling out of a tree to the ground.

(a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel’s velocity just before hitting the ground, assuming it fell from a height of 3.0 m.

(b) If the squirrel stops in a distance of 2.0 cm through bending its limbs, compare its deceleration with that of the airman in the previous problem.

Solution:

Part A

We are given the following: $v_0=0 \ \text{m/s}$ ; $a=-9.80 \ \text{m/s}^2$ ; and $\Delta x=-3.0\ \text{m}$ .

Note that the acceleration is due to gravity and its value is constant at $a=-9.80 \ \text{m/s}^2$ . Also, the distance $x$ , is negative because of the direction of motion. For free-fall, downward motion is considered negative. To solve for the velocity just before it hits the ground, we will solve $v_f$ in the formula

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

Substituting the given values:

\begin{align*} \left(v_f\right)^2 & =\left(v_0\right)^2+2a\Delta x \\ v_f & =\sqrt{\left(v_0\right)^2+2a\Delta x} \\ v_f & = \sqrt{\left(0\:\text{m/s}\right)^2+2\left(-9.80\:\text{m/s}^2\right)\left(-3.0\:\text{m}\right)} \\ v_f & = 7.7\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We are given the following: $v_f=0 \ \text{m/s}$ ; $v_0=7.7 \ \text{m/s}$ ; and $\Delta x=0.02 \ \text{m}$

To solve for the acceleration we shall use the same formula as that in Part A. Solving for the acceleration $a$ :

a=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substituting the given values:

\begin{align*} a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\ a & =\frac{\left(0\:\text{m/s}\right)^2-\left(7.7\:\text{m/s}\right)^2}{2\left(0.02\:\text{m}\right)} \\ a & =-1.5\times 10^3\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

This is approximately 3 times the deceleration of the airman from the previous problem, who was falling from thousands of meters high.

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College Physics by Openstax Chapter 2 Problem 34

In World War II, there were several reported cases of airmen who jumped from their flaming airplanes with no parachute to escape certain death. Some fell about 20,000 feet (6000 m), and some of them survived, with few life-threatening injuries. For these lucky pilots, the tree branches and snow drifts on the ground allowed their deceleration to be relatively small. If we assume that a pilot’s speed upon impact was 123 mph (54 m/s), then what was his deceleration? Assume that the trees and snow stopped him over a distance of 3.0 m.

Solution:

We are given the following: $x=3\ \text{m}$ ; $v_0=54\ \text{m/s}$ ; and $v_f=0 \ \text{m/s}$ .

We are required to solve for the acceleration, and we are going to use the formula

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

Solving for acceleration $a$ in terms of the other variables:

a=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substituting the given values:

\begin{align*} a & = \frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\ a & =\frac{\left(0\:\text{m/s}\right)^2-\left(54\:\text{m/s}\right)^2}{2\left(3.0\:\text{m}\right)} \\ a & =-486\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The negative acceleration means that the pilot was decelerating at a rate of 486 m/s every second.

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College Physics by Openstax Chapter 2 Problem 33

An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m.

a) What is his deceleration?

b) How long does the collision last?

Solution:

We are given the following: $v_0=7.50\:\text{m/s}$ ; $v_f=0.00\:\text{m/s}$ ; and $\Delta x=0.350\:\text{m}$ .

Part A

We are going to use the formula

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

Solving for the acceleration $a$ in terms of the other variables:

a=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substituting the given values:

\begin{align*} a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\ a & =\frac{\left(0\:\text{m/s}\right)^2-\left(7.50\:\text{m/s}\right)^2}{2\left(0.350\:\text{m}\right)} \\ a & =-80.4\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We are going to use the formula

\Delta x=v_{ave}t

Since $v_{ave}=\frac{v_f+v_0}{2}$ , we can write the formula as

\Delta x=\frac{v_f+v_0}{2}\cdot t

Solving for time $t$ in terms of the other variables:

\:t=\frac{2\Delta x}{v_f+v_0}

Substituting the given values:

\begin{align*} t&=\frac{2\Delta x}{v_f+v_0} \\ t&=\frac{2\left(0.350\:\text{m}\right)}{0\:\text{m/s}+7.50\:\text{m/s}} \\ t& =9.33\times 10^{-2}\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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College Physics by Openstax Chapter 2 Problem 32

A woodpecker’s brain is specially protected from large decelerations by tendon-like attachments inside the skull. While pecking on a tree, the woodpecker’s head comes to a stop from an initial velocity of 0.600 m/s in a distance of only 2.00 mm.

a) Find the acceleration in m/s² and in multiples of g (g=9.80 m/s²).

b) Calculate the stopping time.

c) The tendons cradling the brain stretch, making its stopping distance 4.50 mm (greater than the head and, hence, less deceleration of the brain). What is the brain’s deceleration, expressed in multiples of g?

Solution:

We are given the following values: $v_0=0.600 \ \text{m/s}$ ; $v_f=0.000\:\text{m/s}$ ; and $\Delta x=0.002\:\text{m}$ .

Part A

The acceleration is computed based on the formula,

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

Solving for acceleration $a$ in terms of the other variables, we have

a=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substituting the given values,

\begin{align*} a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\ a & =\frac{\left(0.000\:\text{m/s}\right)^2-\left(0.600\:\text{m/s}\right)^2}{2\left(0.002\:\text{m}\right)} \\ a & =-90.0\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

In terms of $g$ taking absolute values of the acceleration , we have

\begin{align*} a & = 90.0 \ \text{m/s}^2 \cdot \left( \frac{g}{9.80 \ \text{m/s}^2} \right) \\ a & = \frac{90.0}{9.80}g \\ a & = 9.18g \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We shall use the formula

\Delta x=v_{ave}t

where $v_{ave}$ is the average velocity computed as

v_{ave}=\frac{v_0+v_f}{2}

Solving for time $t$ in terms of the other variables, we have

t=\frac{2\Delta x}{v_0+v_f}

Substituting the given values, we have

\begin{align*} t & =\frac{2\Delta x}{v_0+v_f} \\ t & =\frac{2\left(0.002\:\text{m}\right)}{0.600\:\text{m/s}+0.000\:\text{m/s}} \\ t & =6.67\times 10^{-3}\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

Employing the same formula we used in Part A, we have

\begin{align*} a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\ a & =\frac{\left(0.000\:\text{m/s}\right)^2-\left(0.600\:\text{m/s}\right)^2}{2\left(0.0045\:\text{m}\right)} \\ a & =-40.0\:\text{m/s}^2 \end{align*}

In terms of $g$ , taking absolute values of the acceleration

\begin{align*} a & = 40.0 \ \text{m/s}^2 \cdot \left( \frac{g}{9.80 \ \text{m/s}^2} \right) \\ a & =\frac{40.0}{9.80}g \\ a & = 4.08g \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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College Physics by Openstax Chapter 2 Problem 31

A swan on a lake gets airborne by flapping its wings and running on top of the water.

(a) If the swan must reach a velocity of 6.00 m/s to take off and it accelerates from rest at an average rate of 0.350 m/s², how far will it travel before becoming airborne?

(b) How long does this take?

Solution:

Part A

We are given the following values: $v_f=6.00\:\text{m/s}$ ; $v_0=0\:\text{m/s}$ ; and $a=0.350\:\text{m/s}^2$ .

From the kinematic equations, the most applicable formula to solve for the change in distance, $\Delta \text{x}$ , is

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

Solving for $\Delta \text{x}$ in terms of the other variables, we have

\Delta x=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2a}

Substituting the given values,

\begin{align*} \Delta x & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2a} \\ \Delta x & =\frac{\left(6.00\:\text{m/s}\right)^2-\left(0.00\:\text{m/s}\right)^2}{2\left(0.350\:\text{m/s}^2\right)} \\ \Delta x & =51.4286\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

From the formula $v_f=v_0+at$ , solve for time, $t$ in terms of the other variables.

t=\frac{v_f-v_0}{a}

Substitute the given values

\begin{align*} t & =\frac{v_f-v_0}{a} \\ t & =\frac{6.00\:\text{m/s}-0.00\:\text{m/s}}{0.350\:\text{m/s}^2} \\ t & =17.1429\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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College Physics by Openstax Chapter 2 Problem 30

A fireworks shell is accelerated from rest to a velocity of 65.0 m/s over a distance of 0.250 m.

(a) How long did the acceleration last?

(b) Calculate the acceleration.

Solution:

We are given the following values: $v_0=0\:\text{m/s}$ ; $v_f=65.0\:\text{m/s}$ ; and $\Delta x=0.250\:\text{m}$ .

We can immediately solve for the acceleration using the given values, so we are going to answer Part B first.

Part B

Solve for the acceleration first using the formula

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

We solve for acceleration in terms of the other variables.

a=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substitute the given values

\begin{align*} a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\ a & = \frac{\left(65.0\:\text{m/s}\right)^2-\left(0\:\text{m/s}\right)^2}{2\left(0.250\:\text{m}\right)} \\ a & =8450\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part A

To solve for the time of this motion, we shall use the formula

v_f=v_0+at

Solving for time, $t$ , in terms of the other variables we have.

t=\frac{v_f-v_0}{a}

We now substitute the values given, and the computed acceleration to find the time.

\begin{align*} t & =\frac{v_f-v_0}{a} \\ t & =\frac{65.0\:\text{m/s}-0\:\text{m/s}}{8450\:\text{m/s}^2} \\ t & =7.6922\:\times 10^{-3}\:\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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College Physics by Openstax Chapter 2 Problem 29

Freight trains can produce only relatively small accelerations and decelerations.

(a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s² for 8.00 minutes, starting with an initial velocity of 4.00 m/s?

(b) If the train can slow down at a rate of 0.550 m/s², how long will it take to come to a stop from this velocity?

(c) How far will it travel in each case?

Solution:

Part A

We are given the the following: $a=0.0500 \ \text{m/s}^2$ ; $t=8.00 \ \text{mins}$ ; and $v_0=4.00 \ \text{m/s}$ .

The final velocity can be solved using the formula $v_f=v_0+at$ . We substitute the given values.

\begin{align*} v_f& = v_0+at \\ v_f & = 4.00\:\text{m/s}+\left(0.0500\:\text{m/s}^2\right)\left(8.00\:\text{min}\times \frac{60\:\sec }{1\:\min }\right) \\ v_f & = 28.0 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

Rearrange the equation we used in part (a) by solving in terms of $t$ , we have

\begin{align*} t & =\frac{{v_f}-v_0}{a} \\ t & = \frac{0\:\text{m/s}-28\:\text{m/s}}{-0.550\:\text{m/s}^2} \\ t & = 50.91\:\sec\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

The change in position for part (a), $\Delta x$ , or distance traveled is computed using the formula $\Delta x=v_0 t+\frac{1}{2} at^2$ .

\begin{align*} \Delta x & =v_0 t+\frac{1}{2} at^2 \\ \Delta x & =\left(4.0\:\text{m/s}\right)\left(480\:\text{s}\right)+\frac{1}{2}\left(0.0500\:\text{m/s}^2\right)\left(480\:\text{s}\right)^2 \\ \Delta x & = 7680\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

For the situation in part (b), the distance traveled is computed using the formula $\Delta x=\frac{v_f^2-v_0^2}{2 a}$ .

\begin{align*} \Delta x & =\frac{\left(0\:\text{m/s}\right)^2-\left(28.0\:\text{m/s}\right)^2}{2\left(-0.550\:\text{m/s}^2\right)} \\ \Delta x & =712.73\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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College Physics by Openstax Chapter 2 Problem 28

A powerful motorcycle can accelerate from rest to 26.8 m/s (100 km/h) in only 3.90 s.

(a) What is its average acceleration?

(b) How far does it travel in that time?

Solution:

We are given the following: $v_0=0 \ \text{m/s}$ ; $v_f=26.8 \ \text{m/s}$ ; and $t=3.90\ \text{s}$ .

Part A

The average acceleration of the motorcycle can be solved using the equation $\overline{a}=\frac{\Delta v}{\Delta t}$ . Substitute the given into the equation. That is,

\begin{align*} \overline{a} & =\frac{\Delta v}{\Delta t} \\ \overline{a} & =\frac{26.8\:\text{m/s}-0\:\text{m/s}}{3.90\:\text{s}} \\ \overline{a} & =6.872\:\text{m/s}^2\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The distance traveled is equal to the average velocity multiplied by the time of travel. That is,

\begin{align*} \Delta x & =v_{ave}t\\ \Delta x & =\left(\frac{0\:\text{m/s}+26.8\:\text{m/s}}{2}\right)\left(3.90\:\text{s}\right) \\ \Delta x & =52.26\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}

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College Physics by Openstax Chapter 2 Problem 27

In a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m/s to 40.0 m/s in the same direction. If this shot takes 3.33×10^-2 s, calculate the distance over which the puck accelerates.

Solution:

The best equation that can be used to solve this problem is $\Delta x=v_{ave} t$ . That is,

\begin{align*} \Delta x & = v_{ave} t \\ \Delta x & = \left(\frac{8\:\text{m/s}+40\:\text{m/s}}{2}\right)\left(3.33\times 10^{-2}\:\text{s}\right) \\ \Delta x & = 0.7992\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Therefore, the distance over which the puck accelerates is 0.7992 meters.

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An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of 0.150 m/s2 as it goes through. The station is 210 m long.

(a) How long is the nose of the train in the station?

(b) How fast is it going when the nose leaves the station?

(c) If the train is 130 m long, when does the end of the train leave the station?

(d) What is the velocity of the end of the train as it leaves?

Consider a grey squirrel falling out of a tree to the ground.

(a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel’s velocity just before hitting the ground, assuming it fell from a height of 3.0 m.

(b) If the squirrel stops in a distance of 2.0 cm through bending its limbs, compare its deceleration with that of the airman in the previous problem.

An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m.

a) What is his deceleration?

b) How long does the collision last?

A woodpecker’s brain is specially protected from large decelerations by tendon-like attachments inside the skull. While pecking on a tree, the woodpecker’s head comes to a stop from an initial velocity of 0.600 m/s in a distance of only 2.00 mm.

a) Find the acceleration in m/s2 and in multiples of g (g=9.80 m/s2).

b) Calculate the stopping time.

c) The tendons cradling the brain stretch, making its stopping distance 4.50 mm (greater than the head and, hence, less deceleration of the brain). What is the brain’s deceleration, expressed in multiples of g?

A swan on a lake gets airborne by flapping its wings and running on top of the water.

(a) If the swan must reach a velocity of 6.00 m/s to take off and it accelerates from rest at an average rate of 0.350 m/s2, how far will it travel before becoming airborne?

(b) How long does this take?

Part B

A fireworks shell is accelerated from rest to a velocity of 65.0 m/s over a distance of 0.250 m.

(a) How long did the acceleration last?

(b) Calculate the acceleration.

Freight trains can produce only relatively small accelerations and decelerations.

(a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s2 for 8.00 minutes, starting with an initial velocity of 4.00 m/s?

(b) If the train can slow down at a rate of 0.550 m/s2, how long will it take to come to a stop from this velocity?

(c) How far will it travel in each case?

A powerful motorcycle can accelerate from rest to 26.8 m/s (100 km/h) in only 3.90 s.

(a) What is its average acceleration?

(b) How far does it travel in that time?

In a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m/s to 40.0 m/s in the same direction. If this shot takes 3.33×10-2 s, calculate the distance over which the puck accelerates.

An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of 0.150 m/s² as it goes through. The station is 210 m long.

a) Find the acceleration in m/s² and in multiples of g (g=9.80 m/s²).

(a) If the swan must reach a velocity of 6.00 m/s to take off and it accelerates from rest at an average rate of 0.350 m/s², how far will it travel before becoming airborne?

(a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s² for 8.00 minutes, starting with an initial velocity of 4.00 m/s?

(b) If the train can slow down at a rate of 0.550 m/s², how long will it take to come to a stop from this velocity?

In a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m/s to 40.0 m/s in the same direction. If this shot takes 3.33×10^-2 s, calculate the distance over which the puck accelerates.