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A car drives along a freeway, accelerating according to \(a=5 \sin (\pi t)\), where \(t\) represents time in minutes. Find the velocity at any time \(t\), assuming the car starts with an initial speed of \(60 \mathrm{mph}\).

Short Answer

Expert verified
The velocity of the car at any time \( t \) is \( v(t) = 60 + \frac{5}{\pi}(1 - \cos(\pi t)) \).

Step by step solution

01

Identify the problem

We need to find the velocity of a car at any given time, knowing its acceleration is defined by the function \( a(t) = 5 \sin(\pi t) \) and the initial velocity is 60 mph.
02

Write the expression for acceleration

The given acceleration of the car is \( a(t) = 5 \sin(\pi t) \). This is a function of time, \( t \), which we will integrate to find the velocity.
03

Integrate to find velocity function

The velocity function, \( v(t) \), is found by integrating the acceleration function with respect to \( t \). This involves finding \( v(t) = \int 5 \sin(\pi t) \, dt \).
04

Solve the integral

The integral is \( \int 5 \sin(\pi t) \, dt = -\frac{5}{\pi} \cos(\pi t) + C \). This formula follows from the integration of a sine function.
05

Apply initial condition

We know the initial velocity \( v(0) = 60 \). Using this, substitute \( t = 0 \) into the velocity equation \(-\frac{5}{\pi} \cos(0) + C = 60 \). This simplifies to \(-\frac{5}{\pi} + C = 60 \), solving for \( C \) gives \( C = 60 + \frac{5}{\pi} \).
06

Write the final velocity equation

Substitute \( C \) back into the velocity equation to get \( v(t) = -\frac{5}{\pi} \cos(\pi t) + 60 + \frac{5}{\pi} \). Simplify to \( v(t) = 60 + \frac{5}{\pi}(1 - \cos(\pi t)) \).

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Key Concepts

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

Velocity Function
The concept of a velocity function is central to understanding motion in calculus. In this particular exercise, the velocity function is derived from an acceleration function. The acceleration of the car is given by the trigonometric function \( a(t) = 5 \sin(\pi t) \). Integration of this function with respect to time \( t \) results in the velocity function \( v(t) \).
  • The acceleration function provides the rate of change of the car's speed over time.
  • The resulting velocity function describes the car's speed at any given time \( t \).
To find the velocity function, we integrate the given acceleration. This is essential because integration reverses the process of differentiation, which in physical terms translates changes in speed back into speed itself.
Initial Conditions
Understanding and applying initial conditions is crucial in many calculus problems involving integrals. The initial condition in this exercise is that at \( t=0 \), the car's velocity is 60 mph. This piece of information is necessary to determine the constant of integration \( C \) when finding the velocity function.
  • Initial velocity provides a specific value of the velocity function at a known time.
  • Knowing this helps solve for the unknown constant that appears from the indefinite integral.
By substituting \( t=0 \) and \( v(0)=60 \) into the velocity equation, we can solve for \( C \). This step ensures that the velocity function is not just a general solution, but rather tailored to meet the specific conditions of the problem, accurately describing the motion of the car from the moment it starts.
Trigonometric Integration
Trigonometric integration involves integrating functions that involve trigonometric expressions. In our exercise, we need to integrate \( 5 \sin(\pi t) \) with respect to \( t \).
  • Sine and cosine functions are commonly involved in these types of integrals because of their periodic nature.
  • Integrating functions like \( \sin \) or \( \cos \) involves transformations that leverage their known derivatives.
To integrate \( 5 \sin(\pi t) \), the antiderivative is \( -\frac{5}{\pi} \cos(\pi t) \), a result derived from understanding that the derivative of \( \cos \) is \( -\sin \). This integration results in a term that includes a constant multiple of \( \cos \), plus the integration constant \( C \). Carefully performing this integration step is key to deriving an accurate description of the velocity function from the given acceleration function.

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