# Acceleration and velocity of a car

• skp123
In summary, the conversation discusses the calculation of acceleration, velocity, and position of a 400 kg mine car being hoisted up an incline with a slope angle of 30°. The force in the cable is given by the function F = 3200 t2 [N] where t is time in seconds. The initial velocity of the car is 2 m/s and there is no friction. The equations F=ma and a = (mgsin(30) + 3200t^2)t^2 / 2m are used to calculate the acceleration and velocity at different times. The position after 3 seconds can be determined by integrating again.
skp123

## 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:
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.

## 1. What is the difference between acceleration and velocity?

Acceleration is the rate of change of velocity over time, while velocity is the speed and direction of an object's motion.

## 2. How do you calculate acceleration?

Acceleration can be calculated by dividing the change in velocity by the change in time. The formula for acceleration is a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

## 3. What factors affect the acceleration of a car?

The acceleration of a car is affected by its mass, the amount of force applied to it, and the frictional force between the car and the road.

## 4. How does acceleration affect the speed of a car?

Acceleration directly affects the speed of a car. If the acceleration is positive, the car will speed up, and if it is negative, the car will slow down.

## 5. Can a car have a constant velocity and still be accelerating?

Yes, a car can have a constant velocity and still be accelerating if it is changing direction. This is because acceleration also takes into account changes in direction, not just changes in speed.

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