# Need help - Force required for accelerating vehicle

1. Jun 14, 2015

### Mew

Hello,
I would need some guidance to calculate the force required to accelerate a vehicle up to a certain speed over given amount of time. This might seem simple but I am not sure if I am on the right track anymore. I did some online searching and I couldnt really understand it.

1. The problem statement, all variables and given/known data

The vehicle has a mass of 250 Kg and uses 4 rubber tires which have a rolling coefficient of 0.02.
The vehicle must reach 100 km/h 60 seconds after start from 0 km/h (stillstand) on a even plane.

We will ignore the effect of aerodynamic drag to keep things easy for starters.

2. Relevant equations

Fr=c*m
F=m*a

3. The attempt at a solution

I do not have a solution but I would say:

1) Calculate force required to overcome rolling resistance with given tires.
Fr=c*m
Fr=0.02*250 Kg
Fr=5 N

So the rolling resistance will be 5 N with tires of 0.02 coefficient ?

2) Calculate the force required to accelerate vehicle of 250 Kg to 100 km/h in 60 seconds from standstill on an even plane.
This is where I can't figure out the correct formula to use since I have to take into account the acceleration over a given period of time.

Your help will be greatly appreciated.
Brgds
Steven

2. Jun 14, 2015

### vela

Staff Emeritus
How'd you get an answer in newtons from 0.02*250 kg?

You wrote the equation you need above: F = ma. What does F represent? (Hint: it's not the force you're looking for.) What's the acceleration of the car?

3. Jun 14, 2015

### RyanH42

Here my idea (I need to claim that I am in high school my idea can be wrong)
Thing this way theres four tires each tire will cary quarter mass of car and so resistance force will depend it.5 N is true but I didnt understand thats the "total" resistance or one tire's resistance.The other thing use $x=1/2at^2$ equation it comes from differantiation.If you differantiate both side respect to t you get $dx/dt=at$ , $v=at$.
Make km/h to m/sc then write down the equation then find a.If you interest my idea reply me and I can help you.

4. Jun 15, 2015

### dean barry

You dont give the value for local gravity rate (g), you can use 9.81 (m/s)/s, it wont be far away
Id use SI units, so 100 km/h = 27.778 m/s
Assuming a constant accelerating force value makes the problem simpler and is probably what is implied in the question, its safe also to assume that the weight distribution is equal.
The net rolling resistance force (N) for all the tyres combined will be : 250 * g * 0.02
(you can split the weight if you like then calculate each wheel individually, then add the 4 results, but the answer will be the same)

A good starting point might be calculating the acceleration rate required from the given data (Newtons rules of motion), and then the force required to produce this rate of acceleration (as you stated, F = m*a)

5. Jun 16, 2015

### haruspex

As vela has hinted, that equation is wrong. There's something missing.
Vela's hint, and pointing out the incorrect quoted equation, should be enough as a first response. We should give the OP a chance to supply the right equation.

6. Jun 16, 2015

### Mew

@vela
I am sure that I messed up the Fr=c*m equation. To have a result in Newton I need to account for the gravity of 9.81 m/s^2 present. Wiki calls the equation Fr=c*Fn where Fn is normal force.
I think this should be correct:

Fr = 0.02*(250 kg*9.81 m/s^2)
Fr = 0.02*(2452.5)
Fr = 49.05 N

I found a site that explains the F=m*a equation in an easier way and now I understand that I have to include acceleration in m/s^2 for the equation to work.
The required acceleration should then be calculated with this equation a = delta v / delta t , is this correct ? Your acceleration hint pointed me in the right way.

If this is correct it should be like this:

delta v = 100km/h converted to m/s = 27.77 m/s
delta t = 60 seconds

a = delta v / delta t
a = 27.77 m/s / 60 s
a = 0.4628 m/s^2

Now F=m*a makes sense too.

F = m * a
F = 250 kg * 0.4628 m/s^2
F = 115.70 N

Then the total force required to accelerate this vehicle will be 115.70 N + 49.05 N = 164.75 N

@RyanH42
I think the 5 N for the rolling resistance I calculated were wrong since I did not account for gravity.

@dean barry
Yes, weight distribution should be equal or as equal as possible, buts lets assume it is equal for the sake of simplicity.

7. Jun 16, 2015

### RyanH42

You are right friction force is true its 49.05 N.