Calculating Acceleration Using Velocity Function

  • Thread starter Thread starter Marioqwe
  • Start date Start date
  • Tags Tags
    Acceleration
Click For Summary

Homework Help Overview

The problem involves calculating acceleration from a given velocity function, specifically v = 7.5(1 - 0.04x)^0.3, where velocity is in miles per hour and distance in miles. The goal is to determine the acceleration at t = 0, with x = 0 at that time.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between acceleration and velocity, with some attempting to clarify the correct formula for acceleration. There is confusion regarding the variables used, particularly the use of 'x' for time instead of distance.

Discussion Status

Multiple interpretations of the problem are being explored, particularly concerning the units of 'x' and the correct application of the acceleration formula. Some participants have offered guidance on the definitions and relationships involved, while others express uncertainty about the calculations presented.

Contextual Notes

There is a noted inconsistency in the problem statement regarding the units of 'x', which some participants believe should be in miles rather than hours. This has led to questions about potential typos in the problem or the solution.

Marioqwe
Messages
65
Reaction score
4

Homework Statement


The velocity is given by the relation v = 7.5(1 - 0.04x)^0.3 and v is in mi/h and x in mi. Determine the acceleration in ft/s^2 at t = 0 knowing that x = 0 when t = 0.

Homework Equations


The Attempt at a Solution



At t = 0, v = 7.5 mi/h, x=0
I know that a = v dv/dx

a = v[d(7.5(1 - 0.04x)^0.3)/dx] = (7.5)(7.5)(0.3)(-0.04)(1 - 0.04x)^-0.7

a = -0.675 mi/h * 5280ft/mi * h/3600 * h/3600 = -2.75 * 10^-4 ft/s^2

What am I doing wrong?
 
Last edited:
Physics news on Phys.org
Marioqwe said:

Homework Statement


The velocity is given by the relation v = 7.5(1 - 0.04x)^0.3 and v is in mi/h and x in h. Determine the acceleration in ft/s^2 at t = 0 knowing that x = 0 when t = 0.


Homework Equations





The Attempt at a Solution



At t = 0, v = 7.5 mi/h, x=0
I know that a = v dv/dx
That's incorrect. a = dv/dt, not v*dv/dt. (For some reason you are using x for time. t is nearly always used to represent time.)
Marioqwe said:
a = v[d(7.5(1 - 0.04x)^0.3)/dx] = (7.5)(7.5)(0.3)(-0.04)(1 - 0.04x)^-0.7

a = -0.675 mi/h * 5280ft/mi * h/3600 * h/3600 = -2.75 * 10^-4 ft/s^2

What am I doing wrong?
 
Mark44 said:
That's incorrect. a = dv/dt, not v*dv/dt. (For some reason you are using x for time. t is nearly always used to represent time.)

a IS equal to v*dv/dx=(dx/dt)*(dv/dx)=dv/dt. And I think x is supposed to be distance. In spite of the op saying the 'x is in h'. Otherwise what's the point of saying x=0 at t=0? Actually the final answer looks ok to me. There are a few typos in between.
 
Dick is right; x is supposed to be in mi (miles). I'm sorry for the confusion mark44. However, according to my textbook the answer is -2.75 * 10^-6 ft/s^2.
 
Dick said:
a IS equal to v*dv/dx=(dx/dt)*(dv/dx)=dv/dt.
Right. What threw me off was the OP saying that x was in units of hours.
Dick said:
And I think x is supposed to be distance. In spite of the op saying the 'x is in h'. Otherwise what's the point of saying x=0 at t=0? Actually the final answer looks ok to me. There are a few typos in between.
 
Marioqwe said:
Dick is right; x is supposed to be in mi (miles). I'm sorry for the confusion mark44. However, according to my textbook the answer is -2.75 * 10^-6 ft/s^2.

Well, I just checked it again and I still get -2.75*10^(-4) ft/s^2. Typo in the problem or the solution?
 
No typo in the problem so it must be the solution. Thanks for your help.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
6K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
9
Views
2K