Problem 6 Suppose the position function of a particle moving in one dimension is given by a(t) = 5+ 3t + 212 – 0.5t (1.30) where the coefficients are such that the result will be in meters if you enter the time in seconds. What is the particle's velocity at t = 2s? There are two ways you can do this: • If you know calculus, calculate the derivative of (1.30) and evaluate it at t= 2 s. • If you do not yet know how to take derivatives, calculate the limit in the definition (1.8). That is to say, calculate Ar/At with t; = 2s and At equal, first, to 0.1s, then to 0.01 s, and then to 0.001 s. You will need to keep more than the usual 4 decimals in the intermediate calculations if you want an accurate result, but you should still report only 3 significant digits in the final result.

Problem 6 Suppose the position function of a particle moving in one dimension is given by a(t) = 5+ 3t + 212 – 0.5t (1.30) where the coefficients are such that the result will be in meters if you enter the time in seconds. What is the particle's velocity at t = 2s? There are two ways you can do this: • If you know calculus, calculate the derivative of (1.30) and evaluate it at t= 2 s. • If you do not yet know how to take derivatives, calculate the limit in the definition (1.8). That is to say, calculate Ar/At with t; = 2s and At equal, first, to 0.1s, then to 0.01 s, and then to 0.001 s. You will need to keep more than the usual 4 decimals in the intermediate calculations if you want an accurate result, but you should still report only 3 significant digits in the final result.

Name: Problem 6Suppose the position function of a particle moving in one di
Uploaded: 2024-03-05T11:56:34.000Z
Channel: Bartleby
Description: Problem 6Suppose the position function of a particle moving in one dimension is given bya(t) = 5+ 3t + 2t2 – 0.5t3(1.30)where the coefficients are such that the result will be in meters if you enter the time in seconds.What is the particle's velocit

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter2: Motion In One Dimension

Section: Chapter Questions

Problem 24P: In the particle under constant acceleration model, we identify the variables and parameters vxi,...

See similar textbooks

Similar questions

The motion of a particle is given by the parametric equations x=t2-3t y=t2-2t Does the particle ever come to a stop?if so,when and where? Is the particle ever moving straight up and down?if so,when and where? Is the particle ever moving straight horizontally right or left?if so, when and where?
The position of a particle at time t (s) is s(t) = √(t2 + l)m. Compute its average velocity over [2, 5] and estimate its instantaneous velocity at t = 2.
A particle moves along the x-axis so that it’s position at any time t≥0t≥0 is given by x(t)=t3−3t2+t+1x(t)=t3−3t2+t+1. For what values of t, 0≤t≤30≤t≤3 , is the particle’s instantaneous velocity the same as its average velocity on the closed interval [0, 3]? Your solution should clearly show all steps leading to the answer as well as justification for it.
The position of a particle along a coordinate axis at time t (in seconds) is given by s(t)=3t^2−t (in meters). Find the function that describes its acceleration at time t. Provide your answer below: a(t)= m/s^2
In the particle under constant acceleration model, we identify the variables and parameters vxi, vxf, ax, t, and xf xi. Of the equations in Table 2.2, the first does not involve xf xi, the second does not contain ax, the third omits vxf, and the last leaves out t. So, to complete the set, there should be an equation not involving vxi. (a) Derive it from the others. (b) Use the equation in part (a) to solve Problem 23 in one step.
When given a particle’s position along the x-axis is described by the function x(t) = A t + B t2, where t is in seconds, x is in meters, and the constants A and B are given below. By given the variables belowA = -3.6 m/sB = 3.5 m/s2 What would the expressions be, in terms of A, B, and t for the velocity of the particle as a function of time??My solution at first was 3.6 * 7.0 t.
The velocity function of a particle moving along a horizontal line is given by v(t) = t2-8t+15, where t >= 0 is in seconds. The particle is 5/3 units to the left of the origin at the second instant when it changes direction. 1. Determine th etime interval (s) on which the particle is both moving to the right and slowing down. 2. Find the position function of the particle.
III. The velocity-time graph of a particle in one dimension is shown in the figure. Assuming that the particle started at x0 = 0 m, what are the instantaneous position x(t) and velocity v(t) equations (in the form compliant with SI units) that correspond to the interval - from A to B? - from B to D? - from D to E? Roughly sketch the corresponding a-t and p-t ploots for the given v-t plot.
A particle moves along a curve whose parametric equations are: x=e^(-t), y=2cos(3t) and Z=2sin(3t) Determine: velocity and acceleration as a function of time
A particle moves along the yy axis so that its position at any time t, for 0≤t≤5t, is given by y(t)=t4−18t2. Find the intervals in which the particle is speeding up.
answer neatly The acceleration of a particle is given by a = 3t - 2. a. Find the initial velocity such that the particle will have the same position at t=4s and t=0s. b. Compute the velocity at t-4s.
I need to solve v(sub y)=v(sub 0)sin θ(sub 0)-gt to find the time that it takes the projectile to reach maximum height (so I believe I'm solving for t?). I have previously found v(sub 0y) in part (a), but I'm unsure what v(sub 0) and v(sub y) are. How do I find those variables?