Explanation Given: The acceleration and angle are shown in Figure (1). The equation for the variable x is ( 2 t − t 2 ) m . The equation for the variable y is ( t 2 ) m . The time is 2 s . Write the equation of motion of the particle. x = ( 2 t − t 2 ) m Write the equation of motion of the particle. y = ( t 2 ) m Write the expression of position of the particle in vector form. r = { ( 2 t + t 2 ) i + ( t 2 ) j } m (I). Write the expression to calculate the velocity ( v ) . v = d r d t d r = v d t (II). Here, the velocity in vector form is v and rate of change of position of the particle with respect to time is d r d t . Write the expression to calculate the acceleration ( a ) . a = d v d t d v = a d t (III). Here, the acceleration in vector form is a and rate of change of velocity with respect to time is d v d t . Write the expression for the magnitude of relative velocity υ B / A . υ = υ x 2 + υ y 2 (IV). Here, the magnitude of relative velocity is υ B / A and the velocity components is υ x and υ y Write the expression for the direction of components θ = tan − 1 υ y υ x (V). Here, the direction of the components is θ and the velocity components is υ x and υ y . Write the expression for the magnitude of relative acceleration a B / A . a = a x 2 + a y 2 (VI). Here, the magnitude of relative acceleration is a B / A and the acceleration components is a x and a y Write the expression for the direction of components ϕ = tan − 1 a y a x (VII). Here, the direction of the components is θ and the acceleration components is a x and a y From the figure (1) write the expression for difference between the direction angle of velocity and acceleration. α = θ − ϕ (VIII). From the figure (1) the tangential acceleration is expressed as. a t = a cos δ (IX). From the figure (1) the normal acceleration is expressed as. a n = a cos δ (X). Conclusion: Substitute { ( 2 t + t 2 ) i + ( t 2 ) j } m for r in Equation (II). v = d d t { ( 2 t + t 2 ) i + ( t 2 ) j } m = ( 2 + 2 t ) i + ( 2 t ) j m/s (XI). Substitute 1 s for t in Equation (XI). v = ( 2 + 2 t ) i + ( 2 t ) j m/s = ( 2 + 2 ( 2 ) ) i + ( 2 ( 2 ) ) j m/s = { 6 i + 4 j } m/s (XII)

14th Edition

ISBN: 9780133976588

Author: HIBBELER

Publisher: PEARSON CO

See similar textbooks

Concept explainers

Conservation of Energy

In physics, the rule of the conservation of energy states that energy doesn’t disappear from the atmosphere; instead, it changes its form from one type to another. Energy is one of the most important physical quantities in nature, and it cannot be create…

Videos

Question

Chapter 12.7, Problem 153P

To determine

The velocity, normal and tangential components of the particle’s velocity and acceleration when t=2s

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 12.7, Problem 152P

Chapter 12.7, Problem 154P

Students have asked these similar questions

-6- 8 من 8 Mechanical vibration HW-prob-1 lecture 8 By: Lecturer Mohammed O. attea The 8-lb body is released from rest a distance xo to the right of the equilibrium position. Determine the displacement x as a function of time t, where t = 0 is the time of release. c=2.5 lb-sec/ft wwwww k-3 lb/in. 8 lb Prob. -2 Find the value of (c) if the system is critically damping. Prob-3 Find Meq and Ceq at point B, Drive eq. of motion for the system below. Ш H -7~ + 目 T T & T тт +

Q For the following plan of building foundation, Determine immediate settlement at points (A) and (B) knowing that: E,-25MPa, u=0.3, Depth of foundation (D) =1m, Depth of layer below base level of foundation (H)=10m. 3m 2m 100kPa A 2m 150kPa 5m 200kPa B

W PE 2 43 R² 80 + 10 + kr³ Ø8=0 +0 R²+J+ kr200 R² + J-) + k r² = 0 kr20 kr20 8+ W₁ = = 0 R²+1) R²+J+) 4 lec 8.pdf Mechanical vibration lecture 6 By: Lecturer Mohammed C. Attea HW1 (Energy method) Find equation of motion and natural frequency for the system shown in fig. by energy method. m. Jo 000 HW2// For the system Fig below find 1-F.B.D 2Eq.of motion 8 wn 4-0 (1) -5- m

Chapter 12 Solutions

EP ENGR.MECH.:DYNAMICS-REV.MOD.MAS.ACC.

Ch. 12.2 - a. If s = (2t3) m, where t is in seconds,...Ch. 12.2 - Initially, the car travels along a straight road...Ch. 12.2 - Determine the time of flight when it returns to...Ch. 12.2 - A particle travels along a straight line with a...Ch. 12.2 - A particle travels along a straight line with a...Ch. 12.2 - Determine the time when the velocity of the...Ch. 12.2 - A particle travels along a straight line with an...Ch. 12.2 - A particle moves along a straight line such that...Ch. 12.2 - Determine the acceleration of the particle at s =...Ch. 12.2 - What is the particles velocity when t = 6 s, and...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

The velocity, normal and tangential components of the particle’s velocity and acceleration when t = 2 s

Solution Summary: The author explains the velocity, normal and tangential components of the particle's velocity and acceleration when time t=2s is 0.555.

Concept explainers

Videos

The velocity, normal and tangential components of the particle’s velocity and acceleration when t=2s

Want to see the full answer?

Chapter 12 Solutions