Acceleration and velocity of a car

AI Thread Summary
The discussion focuses on calculating the acceleration and velocity of a mine car being hoisted up an incline. The force acting on the car is given by F = 3200 t^2, and the initial conditions include a mass of 400 kg and an initial velocity of 2 m/s. The correct approach involves using the equation F = ma, accounting for gravitational force, and integrating to find acceleration and velocity. After correcting initial equations, the calculated acceleration at t = 1 s is 3.1 m/s², and the need to include initial velocity in further calculations is emphasized. The position after 3 seconds requires a different integration approach rather than a simple linear formula.
skp123
Messages
22
Reaction score
0

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 ?
 
Last edited by a moderator:
Physics news on Phys.org
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top