
#1
Oct607, 12:18 AM

PF Gold
P: 820

1. The problem statement, all variables and given/known data
x(t)= (7.00m/s)t + (7.00m/s^2)t^2  (2.00m/s^3)t^3 a) find the expression for the velocity as a function of time. b) plot the graphs of the position, velocity and acceleration as accleration vs time for the interval given. c) At what time(s) btw t=0 and t= 3.00 is the particle at rest? d) for each time calculated in part c) is the acceleration of the particle positive or negative? e) calculate the accelration of the particle when the velcocity of the particle is at maximum. f) does the particle have uniform acclerated motion? explain. 2. Relevant equations x(t)= (7.00m/s)t + (7.00m/s^2)t^2  (2.00m/s^3)t^3 3. The attempt at a solution a) I need to see if my thinking is correct as well as what I did.... v(t)= x'(t)= 7.00m + (14.00m/s)t  (6.00m/s^2)t^2 rearranged:v(t)= (6.00m/s^2)t^2 + (14.00m/s)t + 7.00m b) for plotting the graph of velocity and acceleration vs time it would be the equation for my piece and there would be 2 graphs right? c.) yes I found it to be when the graph v is flat at the top of the parabola..I see that when plug into my calculator and that is when t or x = 1.027 to t or x= 1.311 where v(t)= 15.11 (the flat part = when the accleration = to 0) d.) If the velocity is 0 wouldn't the acceleration be constant and thus would be positive ? For this one I'm not sure ... e.) How would I find the maximum velocity? would that be the max point in the graph? I assume that is also when the acceleration of the particle is 0 since that is the highest point in the v vs t graph but also where the acceleration is 0. f.) I'm confused about this...wouldnt' the acceleration be 0 when the particle's velocity is max?? g.) I say that the particle does not have uniform accelerated motion since it first starts out as a slanted line toward the right and it is accelerating at a constant rate then it levels off and eventually slants down so the acceleration goes down. Is this alright?? I'm not sure about d, e, and f Thank you =D 



#2
Oct607, 12:19 AM

PF Gold
P: 820

Is my derivation correct? I was wondering about that negative..does it go into the bracket or stay on the outside?




#3
Oct607, 12:29 AM

P: 1,757

your units should remain the same
i'm still working out your prob, just replying early. a. everything is fine except your units and the negative should be wherever you like :D idk really. doesn't bother me. b. yes you will need to plot 2 graphs or you can do them on 1 graph which shows the relationship which is more convenient c. that is when velocity is at a max, same thing you said in e. d. take the derivative of your velocity function e. yes f. idk g. idk 



#4
Oct607, 01:21 AM

HW Helper
P: 4,125

Deriving the velocity from a distance equation
for part c), they ask when the object is at rest, hence when the velocity is 0... not when the velocity is at its maximum.




#5
Oct607, 08:01 PM

PF Gold
P: 820

a) Is my derivation that I did correct before?
b)graphed it. c) well looking at the graph assuming my derivation was correct, it goes and hits the 0 line for the equation that I derived at only 1 point from 03 sec time. t= 2.752 while the distance equation graph shows that distance is constant at t= 2.765 However if the question states that at which times is it constant...would that mean that there is more than 1 time ? d)and e) would the acceleration be positive?? I don't know but if the velocity is 0 the acceleration is constant...so since: d= 30.645m at 0 velocity point t= 2.7s v= 0m/s is it a=v/t so...a= 0/2.7s= 0 accleration when v= 0m/s f) I say it doesn't since the graph shows the velocity increasing then decreasing to zero and if it was constant accelerated motion it would just increase or decrease in a straight line instead of being a parabola shape.. Is this fine? Thank you 



#6
Oct607, 08:20 PM

HW Helper
P: 4,125

The velocity should be 0 at exactly the same time that the derivative of distance is 0(ie distance is at a max or a min) 2.752 and 2.765 are close enough... Anyway this has nothing to do with the question about where the object is at rest. BTW if velocity is 0 that does not mean acceleration is constant... Rather if the acceleration is 0... then the derivative of the velocity is 0. 



#7
Oct607, 10:42 PM


#8
Oct807, 01:13 AM

HW Helper
P: 4,125

Hi christina. When you take the derivatives:
x(t)= (7.00m/s)t + (7.00m/s^2)t^2  (2.00m/s^3)t^3 remember... the units in the coefficients don't change when you take the derivative... they are just constants... leave the units the same. v(t) = 7.00m/s + (14.00m/s^2)t  (6.00m/s^3)t^2 so see the units here will all be m/s when t is plugged in (t is in seconds). BTW for part c), you want when v(t) = 0... you can solve this equation to get an exact answer and see if it fits what you got from your graphs... and a(t) = 14.00m/s^2  (12.00m/s^3)t so these units are m/s^2, once t in seconds is plugged in... so acceleration is definitely negative around the time 2.765s... so that's part d)... for part e)... the velocity is at maximum or minimum when its derivative = 0... ie when acceleration = 0... what time do you get when you solve for a(t) = 0... this should fit with the time you see on the graph... for part f)... no this is not uniform acceleration... uniform acceleration is when acceleration is constant (acceleration is a horizontal straight line)... here acceleration is changing with time. 


Register to reply 
Related Discussions  
Deriving the continuity equation from the Dirac equation (Relativistic Quantum)  Advanced Physics Homework  3  
Deriving the phase velocity of a wavefunction  Advanced Physics Homework  0  
Equation relating launch velocity to distance pulled back.  Introductory Physics Homework  2  
Deriving an equation  Precalculus Mathematics Homework  2  
Deriving the Lorentz Velocity Tranformation  Introductory Physics Homework  4 