1. ## Mechanics

A miniature train runs on a track, which has a straight horizontal section AB. The power of the train's engine t seconds after passing the point A is given by (3 + 8t - 3/5t²)kW. The time taken to travel from A to B is 5s.

a) show work done by the engine between A and B is 90 000J.

b) The total mass of the train and its load is 1000kg, and the speed of the train at A is 4 m/s. Given that the work done against resistances between A and B is 13500J, calculate the speed of the train at the point B.

I really have no idea how to do this question

2. Originally Posted by djmccabie
A miniature train runs on a track, which has a straight horizontal section AB. The power of the train's engine t seconds after passing the point A is given by (3 + 8t - 3/5t²)kW. The time taken to travel from A to B is 5s.

a) show work done by the engine between A and B is 90 000J.
"power" is defined as "work done per unit time" or "work divided by time": P= W/t so, for constant P, W= Pt. For non-constant P, integrate P(t)dt.
$\int_0^5 (3+ 8t- (3/5)t^2) dt$

b) The total mass of the train and its load is 1000kg, and the speed of the train at A is 4 m/s. Given that the work done against resistances between A and B is 13500J, calculate the speed of the train at the point B.
You have calculated the work done in problem (a). Subtract the work done against resistance and you have the change in kinetic energy. The original kinetic energy is given by [tex](1/2)mv^2= (1/2)(1000)(4^2)= 8000 J. Add that to the increase in kinetic energy to get final kinetic energy. Then use
E= (1/2)mv^2= 500 v^2
with that energy to solve for v.
I really have no idea how to do this question

3. Originally Posted by djmccabie
A miniature train runs on a track, which has a straight horizontal section AB. The power of the train's engine t seconds after passing the point A is given by (3 + 8t - 3/5t²)kW. The time taken to travel from A to B is 5s.

a) show work done by the engine between A and B is 90 000J.

in kJ, $\textcolor{red}{W = \int_0^5 3 + 8t - \frac{3t^2}{5} \, dt}$

b) The total mass of the train and its load is 1000kg, and the speed of the train at A is 4 m/s. Given that the work done against resistances between A and B is 13500J, calculate the speed of the train at the point B.

strike error ...
.

my mistake in reading the question ...

Given that the work done against resistances between A and B is 13500J, calculate the speed of the train at the point B.

I read that to be work done by the resistive force ... an egregious error. The equation should be ...

$13500 = \frac{1}{2}(1000)(v_f^2 - 16)$

$27 = v_f^2 - 16$

$v_f = \sqrt{43}$ m/s

4. I have the answers i just don't understand them and wanted somebody elses perspective thankyou for your help.

the answer i have for part B is

p/v-resistance = MA

work done = ∫fdx

m=1000kg
v=4 m/s

work done = change in energy
13500 = 1/2mv² - 1/2mv²
13500 = 1/2*1000*v² - 1/2*1000*4²

13500=500v²-16*500 divide by 500

27 = v²-16
v²=43
v= √43 =6.557

So if anybody knows where these letters and numbers come from could you please translate to me

also i dont know why you integrate P in part a :/

5. Originally Posted by djmccabie

also i dont know why you integrate P in part a :/
power is the rate of doing work, i.e. ...

$\frac{dW}{dt} = P$

$dW = P \, dt$

$W = \int P \, dt$