Acceleration and velocity of a car

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SUMMARY

The discussion focuses on the dynamics of a 400 kg mine car being hoisted up a 30° incline using a force defined by F = 3200 t² [N]. The initial velocity is 2 m/s, and the absence of friction simplifies the calculations. The correct acceleration at t = 1 s is determined to be 3.1 m/s², calculated using the equation a = 8t² - 4.9. The velocity at t = 2 s requires integration of acceleration, with the initial velocity accounted for in the final calculation.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Basic knowledge of kinematics and integration
  • Familiarity with trigonometric functions, specifically sine
  • Ability to perform calculus operations, including integration
NEXT STEPS
  • Study the integration of acceleration to find velocity in motion problems
  • Learn about the effects of friction on inclined planes in physics
  • Explore the application of trigonometric functions in force calculations
  • Review the concept of initial conditions in kinematic equations
USEFUL FOR

Physics students, mechanical engineers, and anyone studying dynamics and motion on inclined planes will benefit from this discussion.

skp123
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Homework Statement

The 400 kg mine car is hoisted up the incline with a slope
angle of 30°. The massless cable connects the mine car and
the motor that induces its motion. The force in the cable is
defined by following function F = 3200 t2 [N], where the
time t is in seconds. The mine car has at t = 0 and s = 0 an
initial velocity v1 = 2 m/s. There is no friction between the
mine car and the incline.
a) Determine the acceleration of the mine car for t = 1 s.
b) Calculate the velocity of the mine car when t = 2 s.

Image of the mine car
http://img14.imageshack.us/i/caraws.jpg/" http://img14.imageshack.us/i/caraws.jpg

Homework Equations



F=ma

mgsin(30) + 3200 t^2 = ma - second derivative - here we have acceleration

now we integrate the above and we get

a= (mgsin(30) + 3200t^2)t / m - first derivative - here we have velocity

now we integrate again and we get

a= (mgsin(30) + 3200t^2)t^2 / 2m - here we have the distance

The Attempt at a Solution



Are my equations correct ?
 
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No.

In the first equation the sign of the force due to gravity and the tension in the cable are the same.

Also you neglect the initial velocity of 2m/s.
 
Ok.Right. Here is what i do now.

F=m*a

F-mgsin(30) = m*a

3200t2 - 400*9.8*1/2 = 400*a

a=8t2 -4.9 - that's the acceleration

We need to find the acceleration after 1 second. Therefore a=8-4.9=3.1 -> a=3.1

Now we have to find the velocity after 2 seconds.

a=dv/dt -> dv=a*dt . Now we integrate and we get the velocity after 2 seconds.


Are my equations now correct ?
 
Looks good, but possibly neglects the initial velocity which needs to be entered. Perhaps you planned on adding the constant after the integration which is fine, just wanted to make sure it wasn't forgetten: ie \int dv=V(t) +Vo
 
Yes. Thank you. And one last question. I need to determine the position of the car after 3 seconds. Can I use this formula -> S=S0 + v0t + 1/2at2
 
No that is used for linear acceleration; instead integrate again, So = 0 in the problem.
 

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