/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 A snowboarder starts from rest a... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 1

A snowboarder starts from rest at the top of a double black diamond hill. As she rides down the slope, GPS coordinates are used to determine her displacement as a function of time: \(x=0.5 t^{3}+t^{2}+2 t\) where \(x\) and \(t\) are expressed in feet and seconds, respectively. Determine the position, velocity, and acceleration of the boarder when \(t=5\) seconds.

Short Answer

Expert verified
Position: 97.5 ft, Velocity: 49.5 ft/s, Acceleration: 17 ft/s虏 at t=5 s.

Step by step solution

01

Identify given equation and values

We have the displacement function given as \(x=0.5t^{3}+t^{2}+2t\). We want to find the position, velocity, and acceleration at \(t=5\) seconds.
02

Calculate position at t=5 seconds

Substitute \(t=5\) into the displacement function:\[x = 0.5(5)^3 + (5)^2 + 2(5) \]Calculate:\[x = 0.5(125) + 25 + 10 = 62.5 + 25 + 10 = 97.5 \] So, the position \(x\) is 97.5 feet.
03

Determine velocity function by differentiating displacement function

The velocity \(v\) is the first derivative of the displacement function. Differentiate \(x=0.5t^3 + t^2 + 2t\):\[v(t) = \frac{d}{dt}(0.5t^3 + t^2 + 2t) = 1.5t^2 + 2t + 2 \]
04

Calculate velocity at t=5 seconds

Substitute \(t=5\) into the velocity function:\[v = 1.5(5)^2 + 2(5) + 2 \]Calculate:\[v = 1.5(25) + 10 + 2 = 37.5 + 10 + 2 = 49.5 \] So, the velocity \(v\) is 49.5 feet per second.
05

Determine acceleration function by differentiating velocity function

The acceleration \(a\) is the first derivative of the velocity function. Differentiate \(v(t) = 1.5t^2 + 2t + 2\):\[a(t) = \frac{d}{dt}(1.5t^2 + 2t + 2) = 3t + 2 \]
06

Calculate acceleration at t=5 seconds

Substitute \(t=5\) into the acceleration function:\[a = 3(5) + 2 \] Calculate:\[a = 15 + 2 = 17 \] So, the acceleration \(a\) is 17 feet per second squared.

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Key Concepts

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

Displacement
Displacement refers to the change in position of an object from its starting point. In this exercise, the displacement is given as a function of time with the formula \( x = 0.5t^3 + t^2 + 2t \). This shows how the position of the snowboarder changes over time.
To find the snowboarder's position at a specific time, such as \( t=5 \) seconds, we substitute \( t \) with 5 in the equation.
  • Start by calculating \( 0.5(5)^3 \), which results in \( 0.5 imes 125 = 62.5 \).
  • Next, calculate \( 5^2 = 25 \).
  • Add \( 2 imes 5 = 10 \).
Adding these values together, \( 62.5 + 25 + 10 \), we discover that the displacement is 97.5 feet at \( t=5 \) seconds.
Velocity
Velocity represents the rate of change of displacement with respect to time. It tells us how fast the snowboarder's displacement is changing.
To find the velocity function, we take the first derivative of the displacement function \( x = 0.5t^3 + t^2 + 2t \), using the rules of differentiation.
  • The derivative of \( 0.5t^3 \) is \( 1.5t^2 \).
  • The derivative of \( t^2 \) is \( 2t \).
  • The derivative of \( 2t \) is 2.
Combining these results, the velocity function becomes \( v(t) = 1.5t^2 + 2t + 2 \). To determine the velocity at \( t=5 \) seconds, substitute 5 into this equation. Calculating \( 1.5(25) + 10 + 2 \), we find that the velocity is 49.5 feet per second.
Acceleration
Acceleration is the rate of change of velocity with respect to time. It tells us how quickly the velocity of the snowboarder is changing. To find the acceleration function, differentiate the velocity function \( v(t) = 1.5t^2 + 2t + 2 \).
  • The derivative of \( 1.5t^2 \) is \( 3t \).
  • The derivative of \( 2t \) is 2.
  • The derivative of 2 is 0 because it's a constant.
So, the acceleration function is \( a(t) = 3t + 2 \). At \( t=5 \) seconds, substitute 5 into the acceleration function to find \( a = 3(5) + 2 = 17 \). This means the acceleration is 17 feet per second squared.
Differentiation
Differentiation is a mathematical process used to find the rate at which a function is changing at any given point. It is fundamental in calculus and allows us to determine both velocity and acceleration from a displacement function. In differentiation:
  • If you have a term like \( t^n \), its derivative is \( nt^{n-1} \).
  • It's the tool that transforms a displacement function into a velocity function, and a velocity function into an acceleration function.
  • It applies the power rule, sum rule, and constant rule in our calculations.
Differentiation simplifies complex motion into understandable and calculable rates of change.
Calculus
Calculus is a branch of mathematics focusing on rates of change and the accumulation of quantities. Within the realm of kinematics, calculus provides the tools necessary to understand motion in terms of displacement, velocity, and acceleration.
  • It involves differentiation, which helps find derivatives to study changing quantities.
  • In motion analysis, these derivatives give us information about how an object's motion evolves over time.
  • Calculus also helps us apply conceptual understanding to solve real-world problems, like the snowboarder's journey down the hill.
In short, calculus is indispensable for detailed and precise analysis of how objects in motion behave in space and time.

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Most popular questions from this chapter

The diameter of the eye of a stationary hurricane is \(20 \mathrm{mi}\) and the maximum wind speed is \(100 \mathrm{mi} / \mathrm{h}\) at the eye wall with \(r=10 \mathrm{mi}\). Assuming that the wind speed is constant for constant \(r\) and decreases uniformly with increasing \(r\) to \(40 \mathrm{mi} / \mathrm{h}\) at \(r=110 \mathrm{mi}\), determine the magnitude of the acceleration of the air at \((a) r=10 \mathrm{mi},(b) r=60 \mathrm{mi},\) (c) \(r=110 \mathrm{mi} .\)

Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a circular orbit \(100 \mathrm{mi}\) above the surface of the planet. (See information given in Probs. \(11.153-11.154 .)\) $$ \begin{array}{l}{11.155 \text { Venus: } g=29.20 \mathrm{ft} / \mathrm{s}^{2}, R=3761 \mathrm{mi}} \\ {11.156 \mathrm{Mars}: g=12.17 \mathrm{ft} / \mathrm{s}^{2}, R=2102 \mathrm{mi}} \\ {11.157 \text { Jupiter: } g=75.35 \mathrm{ft} / \mathrm{s}^{2}, R=44,432 \mathrm{mi}}\end{array} $$

The acceleration of a particle is defined by the relation \(a=3 e^{-0.2 t},\) where \(a\) and \(t\) are expressed in \(\mathrm{ft} / \mathrm{s}^{2}\) and seconds, respectively. Knowing that \(x=0\) and \(v=0\) at \(t=0,\) determine the velocity and position of the particle when \(t=0.5 \mathrm{s}\).

A projectile enters a resisting medium at \(x=0\) with an initial velocity \(v_{0}=900 \mathrm{ft} / \mathrm{s}\) and travels \(4 \mathrm{in.}\) before coming to rest. Assuming that the velocity of the projectile is defined by the relation \(v=v_{0}-k x\) where \(v\) is expressed in fts and \(x\) is in feet, determine (a) the initial acceleration of the projectile, \((b)\) the time required for the projectile to penetrate 3.9 in. into the resisting medium.

Instruments in an airplane which is in level flight indicate that the velocity relative to the air (airspeed) is \(120 \mathrm{km} / \mathrm{h}\) and the direction of the relative velocity vector (heading) is \(70^{\circ}\) east of north. Instruments on the ground indicate that the velocity of the airplane (ground speed) is \(110 \mathrm{km} / \mathrm{h}\) and the direction of flight (course) is \(60^{\circ}\) east of north. Determine the wind speed and direction.
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