Problem related to power of engine and resistance

[songoku]

Homework Statement

A car of mass 500 kg has a maximum speed of 40 ms^-1 on a level road with the engine of the car working at a constant power of 20 kW. The resistance to motion is proportional to the square of the speed.

(i) find its acceleration when the speed is 20 ms^-1
(ii) find the distance traveled while the speed increases from 20 ms^-1 to 30 ms^-1
(iii) when the car reaches the speed of 30 ms^-1, the power is switched off. Find the time required to reduce the speed from 30 ms^-1 to 20 ms^-1

Homework Equations

P = Fv
F = ma

The Attempt at a Solution

(i)
F = P/v = 20000/20 = 1000 N

a = F/m = 1000/500 = 2 ms^-2

For (ii) and (iii), I don't know...
the resistance = kv², how to find the value of k? and how to continue after that? I think kinematics doesn't work because the acceleration is not constant.

Thanks

 

[Lok]

! the acceleration while a car is drivin at 20 m/s is 0. I think the problem meant the resistance or the force that the engine has to use.

! using simple x=Vo*t+1/2at^2 is easy but a = acc of the car - rezistance (which is a variable) Have fun!

! again v=a*t where a=-rezistance (variable acording to speed) Have fun again! (Btw i'd say t=Infinity, just a wild guess that might be wrong)

 

[rl.bhat]

Hi songoku,
The net force acting on the car is given by
f = F - k*v^2...(1)
When the car reaches its maximum velocity, its acceleration is zero. Hence f = 0.
and F = k*v^2. But F = W/v. So
W/v = k*v^2 or k = W/(vmax)^3. Find k.
Now for (i), find f for given velocity by using eq(1) and hence find acceleration.
For (ii), find f1 and f2 for two given velocities. Then
ΔKE = (f1+f2)/2*s. find s.
For (iii), put W = 0. m*a = -k*v^2 or a = -k/m*v^2
So dv/dt = -(k/m)*v^2 Or
dv/v^2 = -(k/m)*dt. Find the integration to get time t.

 

[songoku]

H rl.bhat and Lok

I think I get it. I'll try it first. Thanks !