Problem 13 - Technical Mathematics with Calculus

Question

The velocity of a moving point is given by the first derivative of the displacement, and the acceleration is given by the second derivative of the displacement. Find the velocity and acceleration at \(t=1.55 \mathrm{s}\), of a point whose displacement is given by $$s=4.55 t^{3}+2.85 t^{2}+5.22 \quad \mathrm{cm}$$ where \(t\) is the elapsed time, in seconds.

Step-by-Step Solution

Answer

The velocity at t = 1.55 s is \(v(1.55) = 13.65(1.55)^2 + 5.7(1.55)\) cm/s and the acceleration at that time is \(a(1.55) = 27.3(1.55) + 5.7\) cm/s².

Step 1: Find the Velocity Function

The velocity, denoted by 'v', is the first derivative of the displacement function 's' with respect to time 't'. Therefore, calculate the first derivative of the displacement equation by applying the power rule: If the displacement is given by the function \(s=4.55t^3 + 2.85t^2 + 5.22\), then \(v(t) = ds/dt = 3(4.55)t^2 + 2(2.85)t^1\). After multiplying the coefficients, the velocity function simplifies to \(v(t) = 13.65t^2 + 5.7t\).

Step 2: Evaluate the Velocity at t = 1.55 s

To find the velocity at a specific time, substitute this time into the velocity function. Using the value of t = 1.55 s in the velocity equation: \(v(t) = 13.65t^2 + 5.7t\), compute \(v(1.55) = 13.65(1.55)^2 + 5.7(1.55)\).

Step 3: Find the Acceleration Function

The acceleration 'a' is the second derivative of the displacement function with respect to time 't'. Hence, differentiate the velocity function to find the acceleration function: Taking the first derivative of \(v(t) = 13.65t^2 + 5.7t\), gives the acceleration equation \(a(t) = dv/dt = 2(13.65)t + 5.7\). After simplifying, the acceleration function is \(a(t) = 27.3t + 5.7\).

Step 4: Evaluate the Acceleration at t = 1.55 s

To determine the acceleration at time t = 1.55 seconds, substitute the time into the acceleration function. This means evaluating \(a(t) = 27.3t + 5.7\) by substituting \(t = 1.55\) to find \(a(1.55) = 27.3(1.55) + 5.7\).

Key Concepts

Velocity and Acceleration

In physics, understanding how objects move is crucial, and this is where concepts like velocity and acceleration come into play. Velocity refers to the rate at which an object changes its position. It's a vector quantity, which means it has both magnitude and direction. Mathematically, we can find an object's velocity by taking the first derivative of its displacement with respect to time.

Acceleration, on the other hand, tells us how quickly an object's velocity changes. It is also a vector quantity. We can determine an object's acceleration by finding the second derivative of the displacement or the first derivative of the velocity, again with respect to time.

Real-World Example:

If you're driving a car and press the gas pedal, the car's speedometer indicates your velocity. If you press down on the pedal harder and the speed increases rapidly, this increase in velocity over time is your acceleration.

First Derivative of Displacement

When studying motion in calculus, the displacement is a function that represents the position of an object relative to a starting point, often given as a function of time, such as s(t). The first derivative of displacement is of profound importance because it provides the object's velocity.

By applying differentiation to the displacement function, we achieve a new function that directly correlates to velocity. To differentiate effectively, it is helpful to understand power rule differentiation, a common method of taking derivatives of polynomial functions.

• For a displacement function s = at^n, where a is a constant and n is a positive integer, the velocity v will be v(t) = nat^(n-1).
• The units of the resulting velocity function inherit the time unit from the original function, typically reflecting a rate of change in position per unit time (e.g., meters per second).

Second Derivative of Displacement

Diving deeper into the motion of an object, the second derivative of displacement with respect to time gives us the object's acceleration. After finding the first derivative to obtain the velocity function, we take the derivative of the velocity function to find acceleration.

The process of finding the second derivative is essentially applying the power rule differentiation again, but this time to the velocity function. The result tells us how the velocity changes over time, which directly relates to acceleration.

Acceleration Function:

Continuing from the example of velocity, if we have a velocity function of the form v = b*t^m, then the acceleration a can be derived by taking the derivative to get a(t) = mb*t^(m-1). This acceleration function is vital for understanding the dynamics of the object's motion.

Power Rule Differentiation

In calculus, the power rule is one of the most straightforward techniques for differentiating expressions where a variable is raised to a power. It's immensely useful in physics to calculate velocity and acceleration.

The rule is simple: for a given function f(x) = x^n, where n is any real number, the derivative f'(x) = nx^(n-1). This method is used in both the first and second derivative cases mentioned earlier.

Applying the Power Rule:

When applying the power rule, bring down the exponent as a coefficient multiplier and reduce the exponent by one. It's an elegant shortcut to finding derivatives without more complicated methods, making it a fundamental tool in solving motion-related problems in calculus.