# Total Distance Travelled

## Homework Statement

Given that the car produced work (W) of 32000 kJ during the trip to move forward on a flat pavement with coefficient of friction of 0.2, how far did the car travel?

Assume that the car was at rest then it accelerates to 100 km/h in 12 s, and then which maintained constant velocity of 100 km/h till gas (work) ran out. The mass of the car is 432.5 kg, and ignore the mass change from consuming gas.

F = ma

## The Attempt at a Solution

I just did this:
W = (ma - umg)d

To find the distance, is this correct?

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dynamicsolo
Homework Helper
First of all, there are two parts of the travel to consider: the first twelve seconds during which the car is accelerating, and the rest of the trip, during which the car maintains a constant speed.

Think about the free body diagram for this vehicle. The net force is ma, but the work the engine is producing involves the applied force (call it F_app). This applied force changes after twelve seconds.

(Something else that tells you that your work equation can't be quite right -- if the car must overcome friction, the term involving friction wouldn't be negative, since this would imply that friction is somehow reducing the required work...)

Regarding the work equation, I figure since car is accerating forward it's +ve and friction is preventing the car from accelerating; thus, -ve.

Isn't this right?

dynamicsolo
Homework Helper
I'm not sure what "ve" is, but the signs are right for the two forces.

So, during the first twelve seconds, we have

F_net = F_app - (mu)·mg = ma ,

while thereafter, we have

F_net' = F_app' - (mu)·mg = 0 .

So the work is done by the car's engine through F_app over the first 12 seconds and through F_app' the remaining time.

Notice that you can figure out how far the car goes in the first 12 seconds, independently of this. That work has to come out of the 32·10^6 J , so the remaining available work will determine the distance traveled the rest of the time. (Don't forget to add the two distances together at the end.)

DaveC426913
Gold Member
I'm not sure what "ve" is...
+ve = positive
-ve = negative

dynamicsolo
Homework Helper
+ve = positive
-ve = negative
Oh-kay, first time I've seen that used... (In context, just writing the signs would have worked.)

DaveC426913
Gold Member
Oh-kay, first time I've seen that used... (In context, just writing the signs would have worked.)
Eliminates ambiguity and heartache. +ve is a property, + is an operator.

So based on what I've understood I derived following equation:
Wnet = Wacc + Wcv

Where:
Wnet = Total work produced by the car. (36000 kJ)
Wacc = Work produced during acceleration.
Wcv = Work produced during constant velocity.

So for acceleration part:
Wacc = (Fapp - Ffric)d
Wacc = (ma - umg)d

We can find the 'd' or distance car travelled during acceleration in above equation by:
d = (Vinit)t + 0.5at^2

For Wcv we have no acceleration, but there's friction force thus:
Wcv = (Ffric)x
Wcv = (umg)x

Where:
x = Distance travelled during constant velocity.

Sub everything and find x, then add d and x together to find how far the car travels:
36000 = (ma - umg)d + (umg)x

--------------------------------------

Another method I found is calculating Wcv energy from energy equation thus:
Wcv = E = 0.5mV^2 (Energy lost due to friction during constant velocity?)

And subsitute to same final equation in the first method and solve for 'd':
36000 = (ma - umg)d + 0.5mV^2

--------------------------------------

Which method is correct?

dynamicsolo
Homework Helper
So for acceleration part:
Wacc = (Fapp - Ffric)d
Keep in mind that it is the engine doing the work through the applied force, so we have F_net = F_app - F_fric and we need

Wacc = F_app · d , with this d given by

d = (Vinit)t + 0.5at^2
with v_init = 0 and t = 12 seconds. (You could even use the "velocity squared" equation to find d:

(v_final)^2 = (v_init)^2 + 2·a·d . )

For Wcv we have no acceleration, but there's friction force thus:
Wcv = (Ffric)x
Wcv = (umg)x

Where:
x = Distance travelled during constant velocity.
This part is fine. We know that W_acc + W_cv = 32·10^6 J , so we can solve for x. The answer to the question, as I'm reading it, is d + x .

Sub everything and find x, then add d and x together to find how far the car travels:
36000 = (ma - umg)d + (umg)x
This will be

32,000,000 = (ma + umg)d + (umg)x .

The total work is 32,000 kilojoules.

Another method I found is calculating Wcv energy from energy equation thus:
Wcv = E = 0.5mV^2 (Energy lost due to friction during constant velocity?)
I'm afraid this isn't so. The kinetic energy is being sustained by the engine at 0.5mV^2 , despite the work done by friction, so this couldn't be the work done by the engine. The engine is working to cancel the effect of friction, so the work done by the engine in the "cruise" phase will be equal in magnitude to the work done by friction, (umg)x .

[It should be mentioned that we are, of course, ignoring air resistance in this problem. In fact, for a real automobile, the effect of air drag is far larger than the effect of road friction.]