# Speed of a Car Question

1. Feb 16, 2009

### TFM

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

A car enthusiast claims to have the “fastest Fiesta in Farnham”. From a standing start, the car covers a quarter of a mile in 10.5 seconds. Making reasonable assumptions, estimate how fast the car is going by the quarter-mile post. Is your calculation likely to be an underestimate or an overestimate?

2. Relevant equations

$$v^2 = u^2 + 2as$$

$$v = u + at$$

3. The attempt at a solution

This seems easy, but both ways I try I find myself thinking too much...

I was first going to go with using:

$$v^2 = u^2 + 2as$$

and inserting a acceleration in, but we aren't given an acceleration (although the question asks to use reasonable assumptions), but I realised that this wouldn't take into account the time of 10.5seconds.

$$v = u + at$$

But get the same problem of not utilising the distance travelled.

What would be the best way to approach this problem. Could it be to rearrange one of the two equations I have noted above, into terms of a (ie a = ...), and then insert this into the other equation, thus utilising both the time and distance?

TFM

2. Feb 16, 2009

### dontdisturbmycircles

The reasonable assumption is that the acceleration is constant, all our lovely equations for constant acceleration are useless if the acceleration is not constant!

Since we are given distances and times, we can use one of our kinematics equations to find the acceleration. $$x=X_o+Vo(t)+1/2a*t^2$$

Can we cancel any of those terms?

3. Feb 16, 2009

### TFM

$$x = X_o + Vo(t) + 1/2a*t^2$$

What is the Large X? is this the initial Speed?

4. Feb 16, 2009

### dontdisturbmycircles

Yup!

5. Feb 16, 2009

### TFM

Well, I can't see anything that cancels

$$x = X_o + Vo(t) + 1/2a*t^2$$

And the thing we don't know is Vo(t) and a.

6. Feb 16, 2009

### dontdisturbmycircles

-From a standing start

7. Feb 16, 2009

### TFM

Well, if the initial velocity = 0, x0 will dissapear...

8. Feb 16, 2009

### Kruum

Just to make things sure the $$x_o$$ is the starting place of the car and $$v_o$$ is the initial velocity.

9. Feb 16, 2009

### dontdisturbmycircles

No it won't! =-) Here is another thing we know

X-Xo = 1/4mile

You should only have one unknown, you know t, you know X-Xo, you know Vo=0, so find a! :)

10. Feb 16, 2009

### dontdisturbmycircles

Yea, thanks Kruum, I kinda assumed that he had come across this equation before but perhaps not, it is one of the three widely used constant acceleration kinematics equations =-).

11. Feb 16, 2009

### TFM

Okay, so:

$$(x - x_0) = 1/4miles$$

$$x - X_o = Vo(t) + 1/2a*t^2$$

$$1/4 = 0 + 1/2a*(10.5)^2$$

1/4 mile = 402m

$$402 = 0 + 1/2a*102$$

$$804 = a*102$$

$$a = 7.9 m/s^2$$

Does this look okay?

(the Kinematic equations I know are:

$$v = u + at$$

$$v^2 = u^2 + 2as$$

$$s = ut + \frac{1}{2}at$$

$$s = \frac{v +u}{2}t$$)

12. Feb 16, 2009

### dontdisturbmycircles

Looks great! So now you know acceleration, now find the speed at the finish line.

Also, don't forget to answer the other question. What do you think about our assumption that the acceleration was constant? Do you think the acceleration increased with time or decreased with time? How would this affect our answer?

13. Feb 16, 2009

### dontdisturbmycircles

Except I get a =7.29m/s^2, probably just a typo on your part though :D.

14. Feb 16, 2009

### TFM

Yeah, that was a small rounding error. I get 7.29m

So now, put values into:

$$v = u + at$$

$$v = 7.29*10.5$$

v = 76.6 m/s = 171m/hr

Very Fast.

Also, for the second part, I say that the acceleration will decrease throughout, not constant, which would mean that this value is a overestimate.

15. Feb 16, 2009

### dontdisturbmycircles

Good! =-)

16. Feb 16, 2009

### TFM

Excellent.

Most appreciated.

TFM

17. Feb 16, 2009

### dontdisturbmycircles

You might want to give a reason for the acceleration decreasing, it is possible that it increases - although I tend to agree with you! Also make sure you converted m/s to km/hr properly, I get 275.76 km/h.

18. Feb 16, 2009

### dontdisturbmycircles

No problem! =-)

19. Feb 16, 2009

### TFM

No, that is 171 miles/hour

As for a reason for the decceleration,

As the car speeds up, the air resistance/friction acting upon it will increase, meaning that the engine will need to do more work to accelrate the car.

Does this sound okay?

20. Feb 16, 2009

### dontdisturbmycircles

Yea, that sounds great =-).