How Do You Calculate Race Car Acceleration and Power Output?

[oyster21]

Homework Statement

Ive been given the following questions to work out but think the answers for the first 2 questions are wrong. can someone please point out where i am going wrong

Relevant Equations

qquad

A new development race car of mass 400 kg accelerates from 20 km/h to 200 km/h over a distance of 120 metres. Frictional and wind resistance forces can be assumed to be 1000N during the acceleration. Determine the following:

(i)The average acceleration from 20 km/h to 200 km/h
(ii)The time taken to accelerate from 20 km/h to 200 km/h
(iii)The tractive force produced by the car to provide this acceleration. Note you must also consider the additional 1000N resistive force
(iv)The car finally reaches a speed of 300km/h. Friction and wind resistance forces are considered to be 2000 N at this speed. Using this force determine the power output that is required to maintain this constant speed?MY ANSWERS:
(i)
a = (v2 - u 2 ) / 2s
= (40,000 - 400) / 240
=165m/s2

(ii)
s=½(u+v)t
=120
=½(20+200)t
t=120/110
t=1.1sec

 

[BvU]

Hello @oyster21 , $\qquad$ !

It looks as if you think 20 km/h is 400 m/s.

You need a relevant equation. Maybe even a few of them. Post the ones you think you will need. Explain the symbols if they are not self-evident

https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

Relevant Equations: $\qquad$ any help or guidance will be muchly appreciated.
Thank you

is not a relevant equation

 

[oyster21]

It looks as if you think 20 km/h is 400 m/s.

You need a relevant equation. Maybe even a few of them. Post then ones you think you will need. Exaplain the symbols if they are not self-evident

https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

is not a relevant equation

thank you

20km/h is 20,000m/s and 200km/h is 200,000m/s

is that rite?

im not very confident when it comes to maths so i feel like the questions above are way beyond me.
but i like a challenge and will stick withthis until i have worked it out.

 

[berkeman]

20km/h is 20,000m/s and 200km/h is 200,000m/s

To convert units, just keep "multiplying by one":

$$20\frac{km}{hr} \frac{1000m}{1km} \frac{1hr}{3600s} = ?? \frac{m}{s}$$

Can you multiply that out and let us know what you get? Just cancel units if they are in both the numerator and denominator...

 

[berkeman]

BTW, we also have an Insights article about units that should help you out:

https://www.physicsforums.com/insights/make-units-work/

 

[oyster21]

1km=1000m
1h=3600sec

20km=20,000m

20,000m / 3600sec = 5.5m/s

Am I close?

 

[berkeman]

1km=1000m
1h=3600sec

20km=20,000m

20,000m / 3600sec = 5.5m/s

Am I close?

Yes, but to two significant figures, I get 5.6m/s. Now can you do the rest of the problem with confidence? Always carry units along in your calculations -- that is one of the most important lessons I learned in my first semester of undergrad. When the TA mentioned the "multiply by one" trick above in one of my first classes, all of us looked around at each other thinking, "I never knew that! What a trick!"

 

[PeroK]

thank you

20km/h is 20,000m/s and 200km/h is 200,000m/s

is that rite?

im not very confident when it comes to maths so i feel like the questions above are way beyond me.
but i like a challenge and will stick withthis until i have worked it out.

If you are studying engineering you must develop some understanding of what numbers mean. 20km/h is the speed a car drives round a built-up area. That's a low speed limit. 20,000m/s is way, way faster than that. You can't possibly look at a number like that and imagine that's a valid vehicle speed. The longest track race in athletics is the 10,000m, 25 laps of the track. Your car is doing 50 laps of the track every second?

If the maths goes wrong (as it may) you need to use your common sense to sanity check the numbers.

Thinking that a car might be doing 200,000 m/s is a total disconnection between numbers and reality that you need to do something about - especiallly if you are an engineer.

 

[oyster21]

If you are studying engineering you must develop some understanding of what numbers mean. 20km/h is the speed a car drives round a built-up area. That's a low speed limit. 20,000m/s is way, way faster than that. You can't possibly look at a number like that and imagine that's a valid vehicle speed. The longest track race in athletics is the 10,000m, 25 laps of the track. Your car is doing 50 laps of the track every second?

If the maths goes wrong (as it may) you need to use your common sense to sanity check the numbers.

Thinking that a car might be doing 200,000 m/s is a total disconnection between numbers and reality that you need to do something about - especiallly if you are an engineer.

ok thanks for your input.

 

[oyster21]

U=20km/h
V=200km/h
S=120m
a=?

V²=U²+2as

rearranged to:
a=V² - U² + 2s

a=200,000² - 20,000² / 2x120

=165,000,000

can someone please check my answer.

 

[PeroK]

You haven't specified units. In any case, 165,000,000 is not the sort of number you are looking for.

Generally, you must pay attention to units. The SI units are the metre and the second. You must convert velocities to metres per second.

And acceleration must be in metres per second per second.

 

[oyster21]

You haven't specified units. In any case, 165,000,000 is not the sort of number you are looking for.

Generally, you must pay attention to units. The SI units are the metre and the second. You must convert velocities to metres per second.

And acceleration must be in metres per second per second.

Am I using the correct equation for this?

 

[PeroK]

Am I using the correct equation for this?

Yes. $v^2 - u^2 = 2as$ is the one you want.

You need to convert the speeds to SI units.

 

[oyster21]

Yes. $v^2 - u^2 = 2as$ is the one you want.

You need to convert the speeds to SI units.

Please can you give me an example?

 

[PeroK]

Please can you give me an example?

$1m/s = 3.6 km/h$

 

[oyster21]

$1m/s = 3.6 km/h$

20,000m / 3600sec = 5.6m/s
200,000m/3600sec = 55.6m/s
55.6 - 5.6 = 50m/s

 

[PeroK]

20,000m / 3600sec = 5.6m/s
200,000m/3600sec = 55.6m/s
55.6 - 5.6 = 50m/s

Okay.

 

[oyster21]

V²= U²+2as

rearranged as:

a= (V² - U²) /2s

20km/h = 20,000m
200km/h = 200,000m
2x120m = 240m

20,000m / 3600sec = 5.6m/s
200,000m / 3600sec =55.6m/s

a= 55.6m/s² - 5.6m/s² / 240m = 12.75m/s

Please can anyone tell me if my answer is correct?

 

[PeroK]

V²= U²+2as

rearranged as:

a= (V² - U²) /2s

20km/h = 20,000m
200km/h = 200,000m
2x120m = 240m

20,000m / 3600sec = 5.6m/s
200,000m / 3600sec =55.6m/s

a= 55.6m/s² - 5.6m/s² / 240m = 12.75m/s

Please can anyone tell me if my answer is correct?

You got the right number, but acceleration is measured in $m/s^2$.

 

[oyster21]

You got the right number, but acceleration is measured in $m/s^2$.

So it should be 12.75m/s²

correct?

 

[PeroK]

So it should be 12.75m/s²

correct?

Yes. One could quibble about the number of decimal places, but that's perhaps not important enough at this stage.