Question

Acceleration The velocity of an automobile starting from rest is $$ v(t)=\frac{100 t}{2 t+15} $$ where \(v\) is measured in feet per second. Find the acceleration at (a) 5 seconds, (b) 10 seconds, and (c) 20 seconds.

Short answer

The acceleration of the automobile at 5, 10 and 20 seconds are 3.75, 2.4 and 1.22 feet per second per second respectively.

Step-by-step solution

Find the derivative of the velocity function

The derivative of the velocity function \(v(t)=\frac{100 t}{2 t+15}\) is the acceleration function. Using the quotient rule for derivatives that states if we have a function in the form u/v then its derivative is \(\frac{u'v-uv'}{v^2}\), where ' denotes the derivative. So following this rule, the derivative is: \(a(t)=\frac{(100)(2t+15) - (100t)(2)}{(2t+15)^2}\). Simplifying this gives us: \(a(t)=\frac{1500}{(2t+15)^2}\).\n

Calculate the acceleration at t=5 seconds

Substituting \(t=5\) into our acceleration function gives us: \(a(5)=\frac{1500}{(2(5)+15)^2} = \frac{1500}{400}\) or \(a(5)=3.75\) feet per second per second.

Calculate the acceleration at t=10 seconds

Substituting \(t=10\) into our acceleration function gives us: \(a(10)=\frac{1500}{(2(10)+15)^2} = \frac{1500}{625}\) or \(a(10)=2.4\) feet per second per second.

Calculate the acceleration at t=20 seconds

Substituting \(t=20\) into our acceleration function gives us: \(a(20)=\frac{1500}{(2(20)+15)^2} = \frac{1500}{1225}\) or \(a(20)=1.22\) feet per second per second.

Key concepts

Understanding Velocity

Velocity is an essential concept in physics and mathematics that measures how quickly an object is moving in a specific direction. When an automobile starts from rest, its velocity increases from zero as time progresses.
In this exercise, the velocity function is given by the equation \(v(t) = \frac{100t}{2t + 15}\), where \(t\) represents time in seconds, and \(v(t)\) is measured in feet per second.

• The numerator \(100t\) indicates a linear increase in speed with time, which typically suggests acceleration.
• The denominator \(2t + 15\) accounts for other factors slowing down acceleration, like friction or air resistance.

Understanding how velocity changes over time helps find acceleration, which is the rate of change of velocity.

The Concept of Derivative

A derivative in calculus represents the rate of change of a function with respect to a variable. When applied to a velocity function, the derivative gives us the acceleration of the object.
The velocity of an automobile with respect to time is expressed as \(v(t) = \frac{100t}{2t + 15}\). To find the acceleration, we need the derivative of this function with respect to time \(t\).
This derivative tells us how the velocity is changing over each instant of time. The derivative is essential for understanding how quickly or slowly an object is speeding up or slowing down, known as acceleration.

Applying the Quotient Rule

To find the derivative of a function expressed as a fraction, like our velocity function, we use the quotient rule. The quotient rule is a technique in calculus used to differentiate expressions that are the division of two functions.
The rule states that for functions \(u(t)\) and \(v(t)\), \(\frac{d}{dt}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}\), where \(u'\) and \(v'\) are the derivatives of \(u\) and \(v\) respectively.
Applied to our problem:

• Here, \(u = 100t\) and \(v = 2t + 15\).
• The derivative \(a(t)\) becomes \(\frac{(100)(2t + 15) - (100t)(2)}{(2t + 15)^2}\).
• Simplifying this gives the acceleration function: \(a(t) = \frac{1500}{(2t + 15)^2}\).

Understanding this step is crucial because it allows us to calculate acceleration at different moments in time.

Mastering Calculus Problem-Solving

Successfully solving calculus problems involves a systematic approach to breaking down the problem into manageable parts.
The steps we follow include finding the right mathematical representation, applying calculus rules like the quotient rule, and simplifying the result to a usable formula.
Here's a quick recap of our solution process:

• First, identify the function and what needs to be derived—in our case, the velocity function \(v(t) = \frac{100t}{2t + 15}\).
• Next, use the appropriate calculus rules—here, the quotient rule—to determine the derivative, which represents acceleration.
• Simplify the derivative to get the acceleration function: \(a(t) = \frac{1500}{(2t + 15)^2}\).
• Finally, plug in the specific times \(t = 5, 10, 20\) to calculate the acceleration at these moments: \(3.75\), \(2.4\), and \(1.22\) feet per second squared, respectively.

Approaching problems this way helps demystify the steps and builds confidence in solving similar calculus problems in the future.