# 0-125 kilometer per hour

1. Jun 8, 2009

### hongwong

Hi All,

I am new here and trying to solve a simple problem.

If a car can go from 0-100 in 7.4 sec, how many meters will it take to reach 125 meters?

Do I need more information to work this out? If someone could help me solve this it would be appreciated as I am trying to get my head around it.
:) hong !

2. Jun 8, 2009

### Pengwuino

Welcome to PF!

Do you mean how many seconds it will take to reach 125 meters? If the acceleration is constant, we know it's equal to the change in velocity divided by the change in time which is obviously 100km/hour divided by 7.4seconds. However, you need the 100km/hour in meters/second so you multiply by 1000m/1km and 1hour/3600sec. So your acceleration is 100*1000/(3600*7.4) m/s^2.

Now with kinematic equations, we know that starting from rest, on object under constant acceleration will have a position of $$x = \frac{{at^2 }}{2}$$. You can solve for the time, t, knowing that you're looking for the time it'll take to reach x = 125m!

3. Jun 8, 2009

### hongwong

thanks for your quick update.

Sorry for making myself unclear

The car is starting from rest. I know the car can go from 0-100KPH in 7.4 sec.

what i need to know is "when the car reaches 125 KPH how many meter would it had travelled”.

4. Jun 8, 2009

### Pengwuino

Ah, ok. Well you can quickly determine how much time it will take to reach 125km/hour if you know it reaches 100km/hour in 7.4 seconds. Simple algebra there! Now with that time, use the same equation to solve for x instead knowing what the time is and what the acceleration, a, of the car is (which you can solve from my first post).

5. Jun 8, 2009

### hongwong

thanks for this Pengwuino. Just to know my back ground I am a motor mechanical.

But I do remeber this stuff from High school :).

just working out some statics for my car. -- so i am very very low level when it comes to this but love to work things out!

Ok

For X I get the following

So X = 612 meter (correction edit)

612 - is ther correct ? just want to double check i am doing it right.

now need to work this out for 125KPH.

Cheers !

Last edited: Jun 8, 2009
6. Jun 8, 2009

### hongwong

This did not sound right to me so went back over it.

I went here http://wiki.answers.com/Q/What_is_the_proper_formula_used_to_calculate_acceleration

Could someone be nice enough to double this simple piece of work i am doing ?

a = Δv/Δt = (vfinal - vinitial) / (tfinal - tinitial)

a= (100*1000/3600)/7.4 (as the said by "Pengwuino" - However, you need the 100km/hour in meters/second so you multiply by 1000m/1km and 1hour/3600sec)

a= 3.7538

____

Want to work out distance -

a = (v^2-u^2)/2s

3.7538 = (100*1000/3600)^2/2s

s= 102 meter ? which sound better for my car .... but just need to confirm if i am not doing anything wrong ... long time since i have been back at the books.

where
a=acceleration (m/s2)
v=final velocity (m/s)
u=initial velocity (m/s)
t=time (s)
s=distance (m).

7. Jun 8, 2009

### cepheid

Staff Emeritus
Step 1, solve for acceleration using the given data.

Step 2, solve for the time it takes to reach 125 km/h using the relationship between velocity and acceleration (simple)

Step 3, given that the car takes this amount of time to reach a speed of 125 km/h, use the kinematics equation for constant acceleration quoted to you by pengwuino in post #2 to calculate the distance travelled.

8. Jun 8, 2009

### Bob S

There is a different way of looking at this problem, which leads to a different answer. This ignores both air drag force (worse at higher speeds) and tire slipping (worse at low speeds). This assumes that the engine is constantly running at maximum power output. At low velocities, the acceleration is faster because of gearing and higher torque at the wheels. If the total energy of the vehicle (mv2/2) is linearly proportional to time, this is equivalent to saying that the power input is constant. This leads to the equation
[1] E(t) = (1/2)m v2(t)=C1 t (where C1 = const)
or
[2] v(t) = C2 sqrt(t)
Using 100 km/hr = 27.78 m/sec at 7.4 sec we get C2=10.21
[3] So v(t) = 10.21 sqrt(t) meters/sec.
To get the distance covered in this time we need to integrate v(t)
[4] x(t) = integral0t[10.21 sqrt(t)]dt = (2/3) 10.21 t3/2 = 6.81 t3/2 meters

We can also get the same equation from another derivation:
Suppose the force is F = d(mv)/dt
[5] W = F dx is work
[6] dW/dt = F dx/dt = power = constant
[7] So dW/dt = d(mv)/dt dx/dt = m v dv/dt =constant, or
[8] E(t) = mv2/2 = C1 t (same as eqn [1])

Last edited: Jun 8, 2009