Duplicate Post: Hi Guys, See Below Thread

[Brinkley23]

Homework Statement

Car with mass of 400kg accelerates from 20km/h to 200km/h over a distance of 120m. Friction and wind resistance forces can be assumed to be 1000N during acceleration.

Relevant Equations

4 equations for linear motion to find displacement, initial velocity, final velocity and uniform linear acceleration.

Hi guys see below thread (duplicated by mistake).

Many Thanks

 

[Brinkley23]

Hi,

I have made an attempt to answer the 4 following questions:

Car mass 400kg accelerates from 20-200km/h over a distance of 120m. A resistive force of 1000N is to be assumed during acceleration.

determine:
a) the average acceleration from 20km/h to 200km/h
b) time taken to accelerate from 20km/h to 200km/h
c) the tractive force to provide this acceleration, considering the additional 1000N resistive force.
d) the power output required to maintain constant speed if the car reaches 300km/h with a resistive force of 2000N (i had no idea how to do this)

See attached images for my attempts at each of the above.

Many Thanks

 

[Lnewqban]

It seems to me that your answer to b) is incorrect, as your equation and numbers do not match.
Always check your units at the end.
My answer is 3.93 seconds.

Your answers to a) and c) seem to be correct.
Nevertheless, I would improve the expression to show what is actually happening.
$Average~acceleration=(F_{tractive}-F_{resistive})/m$
My values are $12.73~m/s^2$ and $6092~N$ respectively.

I can't see all your work on d).
Why don't you have any idea?
Power is the product of force and velocity in this case.
If you work your velocity to show in m/s, you wil obtain Joules/second or watts when combined with Newtons.
My values are $166.7~KW$ or $223.5~HP$.

Note how you need a greater tractive force to accelerate than to keep a constant speed.
Note also that the actual acceleration and resistive forces are not constant in real life, as they depend on the square of the speed.
That is the reason for the problem to give you average values for both.

 

[Brinkley23]

Thanks for replying Lnewqban, After spending some more time familiarising myself with the equations I seemed to understand it much better and I too came up with those final values, which confirms I must have done something right and corrected my initial mistakes.

Thanks again!

 

[Lnewqban]

You are welcome.