Question

An acceleration function of an object moving along a straight line is given. Find the change of the object's velocity over the given time interval. $$ a(t)=10 \mathrm{ft} / \mathrm{s}^{2} \text { on }[0,5] $$

Short answer

The change in velocity over the interval is 50 ft/s.

Step-by-step solution

Understand the Problem

The acceleration function, given by \( a(t) = 10 \text{ ft/s}^2 \), represents how quickly the velocity of the object is changing over time. The goal is to determine the change in velocity over the time interval \([0, 5]\).

Setup the Integration

To find the change in velocity from the acceleration function, integrate \(a(t)\) with respect to \(t\) over the interval \([0,5]\). The integral of acceleration gives us the change in velocity: \[\Delta v = \int_{0}^{5} a(t) \, dt = \int_{0}^{5} 10 \, dt\]

Perform the Integration

Compute the integral \(\int_{0}^{5} 10 \, dt\). Since the acceleration is constant, the integration simplifies to: \[\Delta v = \left[10t\right]_{0}^{5} = 10(5) - 10(0)\]

Calculate the Result

Substitute the limits into the integral: \[\Delta v = 50 - 0 = 50 \text{ ft/s}\]

Finalize the Solution

The change in the object's velocity over the interval \([0, 5]\) seconds is \(50 \text{ ft/s}\). This result shows how much the velocity has increased due to the constant acceleration.

Key concepts

Understanding the Acceleration Function

Acceleration is a concept from calculus that describes how the velocity of an object changes over time. It's defined as the rate of change of velocity. In this exercise, the acceleration function is given as a constant value, \( a(t) = 10 \text{ ft/s}^2 \). This means that for every second, the velocity increases by 10 feet per second.

• Acceleration has units of \( \text{ft/s}^2 \), showing it involves a change in velocity over time.
• A constant acceleration implies the velocity changes uniformly.

Understanding the acceleration function is critical, as it sets the foundation for calculating the change in velocity using calculus techniques such as integration. This constant rate of change allows us to predict how motion trajectories behave over time.

Concept of Integration

Integration is a powerful calculus tool used to find the total accumulation of a quantity, such as finding areas under curves or, in this context, determining velocity output from an acceleration function. When you integrate an acceleration function, you essentially calculate the total change in velocity over a specified time interval. Since the acceleration function given is constant, integrating it becomes straightforward. The integral of a constant function \( c \) over an interval \([a, b]\) is given by \( c(b-a) \). This process converts the continuous acceleration input over a span of time into a finite change in velocity.Integration provides us a bridge from a rate of change (acceleration) to a total quantity (velocity change).

Calculating Velocity Change

Velocity change is a direct outcome of integration when dealing with acceleration functions. Here, it is essential to set up your problem correctly to determine how much the velocity of an object changes over a time interval.

• Start with the acceleration function \(a(t) = 10 \text{ ft/s}^2\).
• Integrate over the given time period \([0, 5]\) to find the change in velocity.
• The antiderivative of a constant, such as 10, is simply \(10t\).

By evaluating this antiderivative at our limits, we find the specific amounts of increase or decrease in velocity over the chosen interval. Here, the result is\( 50 \text{ ft/s} \), showing a net increase given the constant acceleration.

Understanding Definite Integrals

A definite integral, in calculus, is used to compute the exact total value of a continuous function over a particular interval. It provides a precise tool to find the accumulation of quantities, such as the change in velocity from an acceleration function.The definite integral is expressed as \( \int_{a}^{b} f(t) \, dt\), indicating the process of summing up slices under the graph of \( f \) from \( a \) to \( b \). This procedure outputs a fixed numerical result corresponding to a physical aspect, like velocity change.

• In this exercise, \( f(t) = 10 \), a constant.
• The limits are \([0, 5]\), detailing the interval of interest.

Implementing the definite integral translates the constant rate of acceleration into a single, accumulated velocity change, simplifying complex motion analysis into a straightforward calculation.