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Chapter 2: Problem 48

If a ball is given a push so that it has an initial velocity of 5 \(\mathrm{m} / \mathrm{s}\) down a certain inclined plane, the distance it has rolled after \(t\) seconds is \(s=5 t+3 t^{2}\) . (a) Find the velocity after 2 \(\mathrm{s}\) . (b) How long does it take for the velocity to reach 35 \(\mathrm{m} / \mathrm{s}\) ?

Short Answer

Expert verified
(a) 17 m/s after 2 seconds. (b) 5 seconds to reach 35 m/s.

Step by step solution

01

Identify the formula for velocity

The distance function is given by \( s = 5t + 3t^2 \). To find velocity, we need the derivative of the distance function with respect to time \( t \). The velocity \( v(t) \) is the derivative of \( s(t) \).
02

Find the derivative

Let's find the derivative of the distance function: \( s = 5t + 3t^2 \). The derivative is calculated as follows:\[ v(t) = \frac{d}{dt}(5t + 3t^2) = 5 + 6t \].
03

Calculate velocity after 2 seconds

Substitute \( t = 2 \) seconds into the velocity function:\[ v(2) = 5 + 6(2) = 5 + 12 = 17 \ \mathrm{m/s} \].
04

Set up the equation to find when velocity is 35 m/s

We want to find \( t \) for which \( v(t) = 35 \ \mathrm{m/s} \). So, set the equation as follows:\[ 5 + 6t = 35 \].
05

Solve for time when velocity reaches 35 m/s

Solve the equation from Step 4:\[ 6t = 35 - 5 \]\[ 6t = 30 \]\[ t = \frac{30}{6} = 5 \ \mathrm{seconds} \].

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Key Concepts

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

Derivative
The derivative is a fundamental concept in calculus. It represents the rate at which a function is changing at any given point. When we talk about the derivative in relation to the distance function, we are often interested in how quickly an object is moving. We call this change in position over time "velocity."
To compute the derivative of a function, you evaluate how the output of the function changes as the input changes slightly. For the given distance function \(s = 5t + 3t^2\), the derivative tells us how the distance \(s\) changes with time \(t\).
  • To find the derivative, apply basic differentiation rules: constants become zero, and each term is differentiated by reducing the power of \(t\) by one and multiplying by the previous exponent.
  • The derivative \(v(t) = \frac{d}{dt}(s)\) gives the velocity function: \(v(t) = 5 + 6t\).
Velocity
Velocity is often described as the speed of an object in a particular direction. It is the derivative of the distance function with respect to time. In other words, velocity tells us how fast the position of an object is changing.
When you are solving calculus problems involving motion, like our distance equation \(s = 5t + 3t^2\), finding the velocity is a common step. You compute it by differentiating the distance function, which gives us the formula \(v(t) = 5 + 6t\).
  • At \(t = 2\) seconds, the velocity is calculated by substituting \(t = 2\) into \(v(t)\): \(v(2) = 5 + 6(2) = 17\) m/s.
  • The velocity of 17 m/s indicates how fast the ball is rolling on the inclined plane after 2 seconds.
Distance Function
A distance function represents how far an object has traveled over time. It is typically an equation that accounts for the initial position and the effects of acceleration. For the ball rolling down an inclined plane, the distance function provided is \(s = 5t + 3t^2\). This equation models the distance based on initial velocity and acceleration.
  • The term \(5t\) corresponds to the initial velocity, since it linearly multiplies time \(t\).
  • The term \(3t^2\) represents acceleration, since it involves \(t^2\), showing that distance increases more rapidly over time.
Using this function, we can derive the velocity function and further analyze the motion. As in our problem, the distance function helps determine how quickly the distance changes at any given point in time.
Time
Time is a critical variable in motion-related calculus problems. It tells us when certain events happen during the motion of an object. In the given exercise, time is the independent variable used to compute derivatives and interpret the motion of the rolling ball.
Time helps us answer questions like "how long before an object reaches a certain speed?" It was used to find at what time the ball reaches a velocity of 35 m/s by solving the equation for \(v(t)\).
  • With the velocity equation \(v(t) = 5 + 6t\), to find when \(v(t) = 35\) m/s, solve \(5 + 6t = 35\).
  • By isolating \(t\), we get \(t = \frac{30}{6} = 5\) seconds, showing it takes 5 seconds for the velocity to reach 35 m/s.
Time allows us to explore the dynamics of motion, linking velocity and distance through calculus.

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Most popular questions from this chapter

Suppose \(f\) is differentiable on \(\mathbb{R}\) and \(\alpha\) is a real number. Let \(F(x)=f\left(x^{\alpha}\right)\) and \(G(x)=[f(x)]^{\alpha} .\) Find expressions for (a) \(F^{\prime}(x)\) and \((b) G^{\prime}(x)\)

Find equations of both the tangent lines to the ellipse \(x^{2}+4 y^{2}=36\) that pass through the point \((12,3) .\)

If a rock is thrown vertically upward from the surface of Mars with velocity 15 \(\mathrm{m} / \mathrm{s}\) , its height after \(t\) seconds is \(h=15 t-1.86 t^{2}\) (a) What is the velocity of the rock after 2 \(\mathrm{s}\) ? (b) What is the velocity of the rock when its height is 25 \(\mathrm{m}\) on its way up? On its way down?

On page 431 of Physics: Calculus, 2 \(\mathrm{d}\) ed. by Eugene Hecht (Pacific Grove, \(\mathrm{CA}, 2000 ),\) in the course of deriving the formula \(T=2 \pi \sqrt{L / g}\) for the period of a pendulum of length \(L,\) the author obtains the equation \(a_{T}=-g \sin \theta\) for the tangential acceleration of the bob of the pendulum. He then says, "for small angles, the value of \(\theta\) in radians is very nearly the value of \(\sin \theta ;\) they differ by less than 2\(\%\) out to about \(20^{\circ} . "\) (a) Verify the linear approximation at 0 for the sine function: \(\sin x \approx x\) (b) Use a graphing device to determine the values of \(x\) for which sin \(x\) and \(x\) differ by less than 2\(\%\) . Then verify Hecht's statement by converting from radians to degrees.

\(47-48=\) Find an equation of the tangent line to the curve at the given point. $$y=\sin (\sin x), \quad(\pi, 0)$$
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