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I'm trying to figure out the top speed of a vehicle and the time it would take to get there.

Information about the vehicle:

  • Mass $m = 200 \, \mathrm{kg} ;$

  • Engine power $= 1000 \, \mathrm{W} ;$

  • Combined forces of friction and air and road resistances $F_{\text{r}} = 40 \, \mathrm{N}$ (ignoring slope).

For the purpose I am calculating the work that needs to be done for values $v$ from $1$ to $100.$ $$ W = 0.5 m v^2 $$

Then I'm calculating the time it would need, knowing the engine's power using $$ P = \frac{W}{t} \qquad \Rightarrow \qquad t = \frac{W}{P} $$

Knowing the time I calculate the acceleration,$$ a = \frac{v}{t} \,.$$

You can see results in the table below, ignore forces because they are just the product of more miscalculation

V, m/s  V, km/h t, s    a, m/s2 W, J        F1, N   F2, N
1.00    3.60    0.10    10.00   100.00      960.00  2000.00
2.00    7.20    0.40    5.00    400.00      460.00  1000.00
3.00    10.80   0.90    3.33    900.00      293.33  666.67
4.00    14.40   1.60    2.50    1600.00     210.00  500.00
5.00    18.00   2.50    2.00    2500.00     160.00  400.00
6.00    21.60   3.60    1.67    3600.00     126.67  333.33
7.00    25.20   4.90    1.43    4900.00     102.86  285.71
8.00    28.80   6.40    1.25    6400.00     85.00   250.00
9.00    32.40   8.10    1.11    8100.00     71.11   222.22
10.00   36.00   10.00   1.00    10000.00    60.00   200.00
11.00   39.60   12.10   0.91    12100.00    50.91   181.82
12.00   43.20   14.40   0.83    14400.00    43.33   166.67
13.00   46.80   16.90   0.77    16900.00    36.92   153.85
14.00   50.40   19.60   0.71    19600.00    31.43   142.86
15.00   54.00   22.50   0.67    22500.00    26.67   133.33
...
25      90      62.5    0.4     62500       0       80

Obviously I'm mistaken somewhere because acceleration never reaches 0 and speed just keeps going up and up.

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closed as off-topic by David Z Oct 12 '18 at 1:29

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When at top speed, the power being expended by the car is given by:$$P= F \times v$$where F is the force being exerted and v is the velocity.

Given the constant retarding force that must be overcome, and the constant power output; we get:$$1000 = 40\times v$$or$$v=25 \text{ m/s}$$

So the maximum velocity of the car is $25 m/s$. That is, the engine could maintain a velocity of $25$ m/s.

Unfortunately, the force available from the constant power output engine decreases as the car's velocity increases from rest towards $25$ m/s, so the final velocity is approached asymptotically as the engine force decreases to $40$ N. IOW, the car never reaches this velocity, but gets as close as you want, if you're willing to wait...

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  • $\begingroup$ It is an extremely simple setup with just a pair of gears reductor in order to increase torque. So basically my calculations were somewhat accurate, determining the top speed (given a constant resistance force, which does not happen in real life, as mentioned). Does this mean that also my time calculation is correct, it would reach 25m/s in about 60 seconds? $\endgroup$ – php_nub_qq Oct 11 '18 at 23:29
  • $\begingroup$ Work is force x distance, and distance is v x t. So when it reaches 25m/s it is pushing 40N in 25m in 1 sec or 1000J per sec is exerted. $\endgroup$ – PhysicsDave Oct 12 '18 at 2:26
  • $\begingroup$ @PhysicsDave but I'm missing something. F = m * a, right? Then in my first calculation a = 10, m=200, so F=2000N, however the supposedly correct force that is moving the vehicle is 1000N which is marked as F1 in the calculations with subtracted Fr. What is causing this difference? $\endgroup$ – php_nub_qq Oct 12 '18 at 2:54
  • $\begingroup$ It's tricky, the velocity is increasing with time and accn is decreasing and it is not linear. I tried to evaluate this by saying at the end of a certain time period the kinetic energy is half x m x v final ^2 , but that energy lost due to the 40N was based on an average velocity ( say one half ). Doing this I get max speed of 50m/s which does not seem correct. $\endgroup$ – PhysicsDave Oct 12 '18 at 15:08

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