If [tex]$a(t)$[/tex] is measured in [tex]$m/(sec)^2$[/tex] (meters per second squared) and [tex]$t$[/tex] is measured in seconds, what are the units of measurement for [tex]$\int_a^b a(t) \, dt$[/tex]?



Answer :

To determine the units of the integral [tex]\(\int_a^b a(t) \, dt\)[/tex] given that [tex]\(a(t)\)[/tex] is measured in meters per second squared [tex]\((m/sec^2)\)[/tex] and [tex]\(t\)[/tex] is measured in seconds [tex]\((sec)\)[/tex], follow these steps:

1. Understand the integral: The integral [tex]\(\int_a^b a(t) \, dt\)[/tex] represents the change in velocity over the time interval [tex]\([a, b]\)[/tex].

2. Units of [tex]\(a(t)\)[/tex]: The acceleration [tex]\(a(t)\)[/tex] is given in units of [tex]\(m/sec^2\)[/tex].

3. Units of [tex]\(dt\)[/tex]: The differential [tex]\(dt\)[/tex] represents an infinitesimally small change in time, measured in seconds [tex]\((sec)\)[/tex].

4. Combine the units: When you integrate [tex]\(a(t)\)[/tex] with respect to time:
[tex]\[ \int_a^b a(t) \, dt \][/tex]
you multiply the units of [tex]\(a(t)\)[/tex] by the units of [tex]\(dt\)[/tex]:
[tex]\[ (m/sec^2) \times (sec) \][/tex]

5. Simplify the units: Performing the multiplication:
[tex]\[ (m/sec^2) \times (sec) = m/sec \][/tex]

The units of the integral [tex]\(\int_a^b a(t) \, dt\)[/tex] are therefore meters per second [tex]\((m/sec)\)[/tex].

This means that the result of this integral operation, which represents the change in velocity over the interval [tex]\([a, b]\)[/tex], is measured in units of meters per second ([tex]\(m/sec\)[/tex]).