A particle moves so that its position as a function of time is $$ \overrightarrow{\mathbf{r}}(t)=A \hat{\mathbf{i}}+B t^{2} \hat{\mathbf{j}}+C t \hat{\mathbf{k}} $$ where \(A=1.0 \mathrm{~m}, B=4.0 \mathrm{~m} / \mathrm{s}^{2}\), and \(C=1.0 \mathrm{~m} / \mathrm{s}\). Write ex- pressions for \((a)\) its velocity and \((b)\) its acceleration as functions of time. \((c)\) What is the shape of the particle's trajectory?

Short Answer

Expert verified
The velocity of the particle is \(0\hat{\mathbf{i}}+ 8.0t\hat{\mathbf{j}}+1.0\hat{\mathbf{k}}\) m/s, the acceleration of the particle is \(0\hat{\mathbf{i}}+ 8.0\hat{\mathbf{j}}+0\hat{\mathbf{k}}\) m/s², and the shape of the particle's trajectory is a parabolic curve.

Step by step solution

01

Calculation of velocity

The velocity of the particle is given by the derivative of the position. Thus, \(\overrightarrow{\mathbf{v}}(t) = d\overrightarrow{\mathbf{r}} / dt = 0\hat{\mathbf{i}}+ (2Bt)\hat{\mathbf{j}}+C\hat{\mathbf{k}} = 0\hat{\mathbf{i}}+ (2(4.0)t)\hat{\mathbf{j}}+(1.0)\hat{\mathbf{k}} = 0\hat{\mathbf{i}}+ 8.0t\hat{\mathbf{j}}+1.0\hat{\mathbf{k}}\)
02

Calculation of acceleration

The acceleration is given by the derivative of the velocity. Thus, \(\overrightarrow{\mathbf{a}}(t) = d\overrightarrow{\mathbf{v}} / dt = 0\hat{\mathbf{i}}+ (2B)\hat{\mathbf{j}}+0\hat{\mathbf{k}} = 0\hat{\mathbf{i}}+ (2(4.0))\hat{\mathbf{j}}+0\hat{\mathbf{k}} = 0\hat{\mathbf{i}}+ 8.0\hat{\mathbf{j}}+0\hat{\mathbf{k}}\)
03

Shape of Trajectory

For the shape of the trajectory, observe that only the y-component (vertical component) of the position depends on time, and in particular, it depends quadratically on time, which is characteristic of a parabolic trajectory. The x-component (horizontal component) and z-component do not change with time, indicating that the particle moves along a line in these directions. Therefore, the shape of the trajectory in 3D space would be a parabolic curve.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A locomotive accelerates a 23 -car train along a level track. Each car has a mass of \(48.6\) metric tons and is subject to a drag force \(f=243 v\), where \(v\) is the speed in \(\mathrm{m} / \mathrm{s}\) and the force \(f\) is in N. At the instant when the speed of the train is \(34.5 \mathrm{~km} / \mathrm{h}\), the acceleration is \(0.182 \mathrm{~m} / \mathrm{s}^{2} .(a)\) Calculate the force exerted by the locomotive on the first car. ( \(b\) ) Suppose that the force found in part \((a)\) is the greatest force the locomotive can exert on the train. What, then, is the steepest grade up which the locomotive can pull the train at \(34.5 \mathrm{~km} / \mathrm{h} ?(1\) metric ton \(=1000 \mathrm{~kg} .\) )

In a baseball game, a batter hits the ball at a height of \(4.60 \mathrm{ft}\) above the ground so that its angle of projection is \(52.0^{\circ}\) to the horizontal. The ball lands in the grandstand, \(39.0 \mathrm{ft}\) up from the bottom; see Fig. 4-38. The grandstand seats slope upward at \(28.0^{\circ}\) with the bottom seats \(358 \mathrm{ft}\) from home plate. Calculate the speed with which the ball left the bat. (Ignore air resistance.)

An \(8.5\) -kg object passes through the origin with a velocity of 42 \(\mathrm{m} / \mathrm{s}\) parallel to the \(x\) axis. It experiences a constant \(19-\mathrm{N}\) force in the direction of the positive \(y\) axis. Calculate \((a)\) the velocity and \((b)\) the position of the particle after \(15 \mathrm{~s}\) have elapsed.

A person walks up a stalled 15 -m-long escalator in \(90 \mathrm{~s}\). When standing on the same escalator, now moving, the person is carried up in \(60 \mathrm{~s}\). How much time would it take that person to walk up the moving escalator? Does the answer depend on the length of the escalator?

In a cathode-ray tube, a beam of electrons is projected horizontally with a speed of \(9.6 \times 10^{8} \mathrm{~cm} / \mathrm{s}\) into the region between a pair of horizontal plates \(2.3 \mathrm{~cm}\) long. An electric field between the plates causes a constant downward acceleration of the electrons of magnitude \(9.4 \times 10^{16} \mathrm{~cm} / \mathrm{s}^{2} .\) Find \((a)\) the time required for the electrons to pass through the plates, \((b)\) the vertical displacement of the beam in passing through the plates, and (c) the horizontal and vertical components of the velocity of the beam as it emerges from the plates.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free