## The position role also indicates direction

A usual application the derivatives is the relationship in between speed, velocity and also acceleration. In this problems, she usually provided a place equation in the type “???x=???” or “???s(t)=???”, which tells you the object’s street from some reference point. This equation likewise accounts because that direction, for this reason the distance could be negative, relying on which direction her object moved away indigenous the recommendation point.

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Average rate of the object is

??? extaverage speed=frac extdistance exttime???

Average velocity the the object is

???frac extchange in position extchange in time=fracs(b)-s(a)b-a???

To find velocity, take it the derivative of your original position equation. Rate is the absolute value of velocity. Velocity accounts because that the direction the movement, so it deserve to be negative. It’s prefer speed, but in a particular direction. Speed, on the other hand, have the right to never be an unfavorable because it no account because that direction, i beg your pardon is why rate is the absolute value of velocity. To uncover acceleration, take it the derivative the velocity.

## Using the position role to uncover velocity and acceleration

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## How to discover velocity and also acceleration

**Example**

Suppose a bit is relocating along the ???x???-axis so that its position at time ???t??? is provided by the formula

???s(t)=3t^2+8t-2t^frac52???

Compute that is velocity and also acceleration as features of ???t???. Next, decision in i beg your pardon direction (left or right) the fragment is relocating when ???t=1??? and whether the velocity and also speed are increasing or decreasing.

To uncover velocity, we take the derivative that the initial position equation.

???v(t)=s"(t)=6t+8-5t^frac32???

To find acceleration, we take the derivative of the velocity function.

???a(t)=v"(t)=s""(t)=6-frac152t^frac12???

To recognize the direction the the bit at ???t=1???, we plug ???1??? right into the velocity function.

???v(1)=6(1)+8-5(1)^frac32???

???v(1)=9???

Because ???v(1)??? is positive, we have the right to conclude the the fragment is moving in the positive direction (toward the right).

To determine whether velocity is raising or decreasing, we plug ???1??? into the acceleration function, because that will give us the price of adjust of velocity, since acceleration is the derivative the velocity.

???a(1)=6-frac152(1)^frac12???

???a(1)=-frac32???

Since acceleration is an adverse at ???t=1???, velocity have to be diminish at the point.

Since the velocity is positive and also decreasing in ~ ???t=1???, that method that speed is likewise decreasing at that point.

## Calculating instantaneous velocity

We usage the term “instantaneous velocity” to explain the velocity of things at a certain instant in time. Offered an equation the models one object’s place over time, ???s(t)???, we deserve to take the derivative to obtain velocity, ???s"(t)=v(t)???. We deserve to then plugin a certain value for time to calculate instantaneous velocity.

**Example**

Given one object’s position equation,

???s(t)=5t^3+2t-8???

where ???s??? is measure in meters and also ???t??? is measure in seconds,

What is the object’s velocity as soon as ???t=0????

What is the object’s velocity once ???t=25????

If the motion begins at 8:45 a.m., what is the velocity at 8:51 a.m.?

First we need to find the derivative the ???s(t)???.

???s"(t)=v(t)=15t^2+2???

To find velocity as soon as ???t=0???, we’ll plug ???t=0??? into ???v(t)???.

???v(0)=15(0)^2+2???

???v(0)=2???

Velocity at ???t=0??? is ???2???m/s.

To discover velocity once ???t=25???, fine plug???t=25???into ???v(t)???.

???v(25)=15(25)^2+2???

???v(25)=9,377???

Velocity in ~ ???t=25???is ???9,377??? m/s.

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8:45 a.m. Is when movement begins, which method that time coincides with ???t=0???. The concern asks because that velocity after ~ ???6??? minutes have actually passed, at 8:51 a.m., yet in our position equation, ???t??? is measured in seconds. We’ll convert minutes right into seconds.