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Chapter 4: Problem 62

What is the magnitude of the acceleration of a sprinter running at \(10 \mathrm{~m} / \mathrm{s}\) when rounding a turn of radius \(25 \mathrm{~m}\) ?

Short Answer

Expert verified
The magnitude of the acceleration is 4 m/s虏.

Step by step solution

01

Understand the Problem

We need to find the magnitude of the centripetal acceleration experienced by a sprinter running at a speed of \(10\,\text{m/s}\) around a turn with a radius of \(25\,\text{m}\). The formula to find centripetal acceleration \(a_c\) is \(a_c = \frac{v^2}{r}\), where \(v\) is the speed and \(r\) is the radius.
02

Substitute the Values

Substitute \(v = 10\,\text{m/s}\) and \(r = 25\,\text{m}\) into the formula \(a_c = \frac{v^2}{r}\). This gives us \(a_c = \frac{10^2}{25}\).
03

Calculate the Acceleration

Calculate \(10^2\) to get \(100\). Then divide \(100\) by \(25\) to find \(a_c = \frac{100}{25} = 4\,\text{m/s}^2\).
04

Conclusion

The magnitude of the sprinter's acceleration is \(4\,\text{m/s}^2\). This is the centripetal acceleration required to maintain the circular motion along the turn.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Circular Motion
Circular motion is a type of movement where an object follows a circular path. In circular paths, the direction of the object's motion constantly changes, even though the speed might remain the same.
One crucial aspect of circular motion is that there is always an inward force needed to keep the object moving in a circle. This force is known as centripetal force.
  • Centripetal force is directed towards the center of the circle.
  • Without this force, the object would move off in a straight line.
The acceleration called centripetal acceleration accompanies this force to change the direction of the object鈥檚 velocity, making circular motion possible. For the sprinter in our problem, the centripetal acceleration allows them to maintain their curve around the track with constant speed.
Physics Problem Solving
Physics problem-solving often involves breaking down problems into manageable steps.
By using established principles and formulas, one can solve complex questions efficiently. Here's how you might tackle a problem involving centripetal acceleration:
  • First, identify what you're asked to find, such as the magnitude of the acceleration.
  • Next, recognize the situation details, like the object's speed and the curve's radius.
  • Apply the relevant formula, in this case, the centripetal acceleration formula, \(a_c = \frac{v^2}{r}\).
  • Substitute known values, perform calculations, and conclude with a clear statement.

    • These steps enable a structured approach, minimizing errors and enhancing clarity.
Kinematics
Kinematics is the area of physics that deals with the motion of objects without considering forces that cause the motion. It focuses on parameters like speed, velocity, and acceleration.
In the context of circular motion, kinematics helps us examine how these parameters change when an object moves along a curved path.
  • The speed remains constant in uniform circular motion, but the velocity (which includes direction) changes.
  • Centripetal acceleration is a key kinematic aspect in circular motion, affecting the change in the object's velocity direction.
Understanding kinematics enables us to accurately describe and predict motion patterns, which we then apply to solve real-world problems like our sprinter's path around a curve.
Acceleration Calculation
When calculating acceleration in a physics problem, it鈥檚 important to identify whether the acceleration is linear or centripetal. In circular motion, centripetal acceleration is calculated using the formula \(a_c = \frac{v^2}{r}\), where \(v\) is the speed and \(r\) is the radius of the circular path.
For our sprinter, substituting the given values into the formula:
  • \(v = 10 \, \text{m/s}\)
  • \(r = 25 \, \text{m}\)
The calculation follows: \[a_c = \frac{10^2}{25} = \frac{100}{25} = 4 \, \text{m/s}^2\]This result, 4 \(\text{m/s}^2\), is the centripetal acceleration required for the sprinter to maintain the curved path smoothly.

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Most popular questions from this chapter

Ship \(A\) is located \(4.0 \mathrm{~km}\) north and \(2.5 \mathrm{~km}\) east of ship \(B\). Ship \(A\) has a velocity of \(22 \mathrm{~km} / \mathrm{h}\) toward the south, and ship \(B\) has a velocity of \(40 \mathrm{~km} / \mathrm{h}\) in a direction \(37^{\circ}\) north of east. (a) What is the velocity of \(A\) relative to \(B\) in unit-vector notation with \(\hat{i}\) toward the east? (b) Write an expression (in terms of \(\hat{\mathrm{i}}\) and \(\hat{\mathrm{j}}\) ) for the position of \(A\) relative to \(B\) as a function of \(t\), where \(t=0\) when the ships are in the positions described above. (c) At what time is the separation between the ships least? (d) What is that least separation?

A woman who can row a boat at \(6.4 \mathrm{~km} / \mathrm{h}\) in still water faces a long, straight river with a width of \(6.4 \mathrm{~km}\) and a current of \(3.2 \mathrm{~km} / \mathrm{h}\). Let i point directly across the river and \(\hat{j}\) point directly downstream. If she rows in a straight line to a point directly opposite her starting position, (a) at what angle to i must she point the boat and (b) how long will she take? (c) How long will she take if, instead, she rows \(3.2 \mathrm{~km}\) down the river and then back to her starting point? (d) How long if she rows \(3.2 \mathrm{~km} u p\) the river and then back to her starting point? (e) At what angle to \(\hat{\mathrm{i}}\) should she point the boat if she wants to cross the river in the shortest possible time? (f) How long is that shortest time?

A particle is in uniform circular motion about the origin of an \(x y\) coordinate system, moving clockwise with a period of \(7.00\) s. At one instant, its position vector (measured from the origin) is \(\vec{r}=(2.00 \mathrm{~m}) \hat{\mathrm{i}}-(3.00 \mathrm{~m}) \hat{\mathrm{j}}\). At that instant, what is its velocity in unit-vector notation?

A particle starts from the origin at \(t=0\) with a velocity of \(8.0 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}\) and moves in the \(x y\) plane with constant acceleration \((4.0 \hat{1}+2.0 \hat{j}) \mathrm{m} / \mathrm{s}^{2}\). When the particle's \(x\) coordinate is \(29 \mathrm{~m}\), what are its (a) \(y\) coordinate and (b) speed?

After flying for 15 min in a wind blowing \(42 \mathrm{~km} / \mathrm{h}\) at an angle of \(20^{\circ}\) south of east, an airplane pilot is over a town that is \(55 \mathrm{~km}\) due north of the starting point. What is the speed of the airplane relative to the air?
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