The position vector r⃗=5.00 t î+(e t+f t^2) ĵ locates a particle as a function of time t. Vector

step 1 This is chapter 4 problem number 10. We're given this graph, the angle versus time. And we're gi

The position vector r⃗=5.00 t î+(e t+f t^2) ĵ locates a...

The position vector $\vec{r}=5.00 t \hat{\mathrm{i}}+\left(e t+f t^{2}\right) \hat{\mathrm{j}}$ locates a particle as a function of time $t$. Vector $\vec{r}$ is in meters, $t$ is in seconds, and factors $e$ and $f$ are constants. Figure 4-22 gives the angle $\theta$ of the particle's direction of travel as a function of $t$ ( $\theta$ is measured from the positive $x$ direction). What are (a) $e$ and (b) $f$, including units?

Key Concepts

Angle of Motion and Component Relationships

The direction of motion of a particle can be described by the angle its velocity vector makes with a reference axis, typically calculated using the inverse tangent function of the ratio of the perpendicular components. This relationship between the components of the velocity and the angle allows one to solve for unknown parameters when additional information about the trajectory’s direction is provided.

Velocity Vector Components

The components of the velocity vector, obtained by differentiating the corresponding components of the position vector, define the velocity in each coordinate direction. Analyzing these components separately is crucial for understanding and solving problems in two-dimensional motion, where the overall motion is a combination of independent motions along each axis.

Differentiation of Vector Functions

Taking the derivative of a position vector with respect to time yields the velocity vector. This process of differentiation is fundamental in physics and calculus because it allows for the determination of the instantaneous rate of change of the position, which in turn describes the object’s motion characteristics such as speed and direction.

Position Vector

A position vector defines the location of an object in space relative to an origin using its coordinate components. In kinematics, the position vector is typically expressed as a function of time to describe the path of the object, and its elements provide the essential starting point for further analysis such as finding velocity and acceleration.

Transcript

00:01 This is chapter 4 problem number 10.

00:03 We're given this graph, the angle versus time.

00:08 And we're given the position vector as 5 times t is a function of time, i -hat, plus some constant e times time, plus some constant f times squared, j -hat.

00:26 And here e and f are constants, and we're asked just that.

00:35 So in part a, we're asked what e is.

00:38 So, well, the derivative, the first derivative of the position vector with respect to time is going to give us the velocity vector, right? so if we take the derivative of the position vector, then we have five.

00:59 I hat plus e plus two times f times t j hat right this is our velocity vector so then we know that the first component is vx right vx here is five and the y here is e plus two f t all right now let's look at the graph and see what's going on let's say at time equals right here at the origin, the corresponding angle would be, so it's going increments of five here in the y direction, 5, 10, 15, 20, 25, 35.

01:44 So when time equals zero, the angle is 35 degrees, right? so then we're going to apply this initial condition to, as you know, in order to figure out the direction of this velocity vector, we can do tangent theta equals the y component divided by the x component of the vector, right? so, y component would be e plus 2ft, just like we wrote here openly, divided by five, right? in five meters per second.

02:20 Okay, let's just keep track of the units as well because that's also what you're asked, okay? five meters per second because now we're, i would take the derivative, so it's meters per second.

02:31 It has to be right, the units for speed.

02:36 Now, we can apply this condition then when times equals zero.

02:40 So wherever you see t plugin zero, it makes this term, entire term, zero, right, to ft, zero.

02:47 So what we have is e over five meters per second and the angle, then tangent 35 degrees.

02:55 So from here, e is going to be equal to 5 meters per second times tangent of 35 degrees, and that is going to give us 3 .5 meters per second.

03:09 So this is what e is.

03:11 Now, in part b, we're asked what f is, the constant f is.

03:16 So for that, let's look at another kind of condition here that is given to us by the graph...

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