# Find the distance between the car and the truck- Mechanics

• chwala
In summary, the car has to travel 900 metres to overtake the truck. The gap between the car and the truck will be 851 - 800.1 = 51.1 metres.
chwala
Gold Member
Homework Statement
See attached
Relevant Equations
##s=vt##
Find the problem and the solution below;

Find my approach to the problem.
Considering the motion of the car;
##v=u+at##
##33.333= 26.6667+30a##
##a=0.2222##
Therefore it follows that,
##s##=## ut##+##\frac {1}{2}####at^2##
##s##=##(26.67 ×30 +(0.5×0.222×30^2)##
##s=900##metres

The distance that car has to cover in order to overtake the truck is given by
##s= 35+10+4=49##metres

Therefore we shall have, ##900-49= 851## metres

Now considering the distance covered by the truck in ##30## seconds, we shall have
##s=vt##
##s=26.67×30=800.1##metres

Therefore the gap between the car and the truck will be ##851-800.1=51.1##metres.

Any other approach highly appreciated.

Last edited:
FactChecker
This is an interesting problem. I would have bet that the rates of acceleration and deceleration would change the answer, even given the restriction that the entire maneuver took half a minute. But I see from their solution that it does not because the horizontal position of point M does not change the area of the triangle AMB.

chwala and hmmm27
It would simplify the arithmetic slightly to work in the rest frame of the truck i.e. subtract ##96 \mathrm{kph}## from all of the velocities.

@chwala, a couple of points.

First, you assume that the acceleration is constant over the half minute (and the deceleration instantaneous). You are not told that. As FactChecker pointed out, it turns out that the rates of acceleration and deceleration don't make a difference, but it's not obvious that you knew that and it's not clear from your working (and it was not obvious to me or FactChecker either). The geometric method, on the other hand, makes it clear.

Second, you once again unnecessarily convert to m/s. your final answer of 51.1 m is slightly erroneous, not because of a mistake in the conversion, but because of premature rounding off. You say
s = 26.67*30 = 800.1 m
Actually it is 26.666666... * 30 = 800 m
The rule is always work to high precision and round off your final answer to the correct number of sig figs at the end. (If you are asked to report an intermediate result, report it to the correct sig figs, but still use the more precise value for subsequent calculations.)

This is regularly going to be an issue when you convert (e.g.) round numbers in km/min or km/h to m/s, because of the factor of 60 or 3600.

Edit: There is also a basic arithmetic error: 851 - 800.1 = 50.9, not 51.1. But still an error due to rounding off.

FactChecker and chwala
mjc123 said:
@chwala, a couple of points.

First, you assume that the acceleration is constant over the half minute (and the deceleration instantaneous). You are not told that. As FactChecker pointed out, it turns out that the rates of acceleration and deceleration don't make a difference, but it's not obvious that you knew that and it's not clear from your working (and it was not obvious to me or FactChecker either). The geometric method, on the other hand, makes it clear.

Second, you once again unnecessarily convert to m/s. your final answer of 51.1 m is slightly erroneous, not because of a mistake in the conversion, but because of premature rounding off. You say
s = 26.67*30 = 800.1 m
Actually it is 26.666666... * 30 = 800 m
The rule is always work to high precision and round off your final answer to the correct number of sig figs at the end. (If you are asked to report an intermediate result, report it to the correct sig figs, but still use the more precise value for subsequent calculations.)

This is regularly going to be an issue when you convert (e.g.) round numbers in km/min or km/h to m/s, because of the factor of 60 or 3600.

Edit: There is also a basic arithmetic error: 851 - 800.1 = 50.9, not 51.1. But still an error due to rounding off.
Second, you once again unnecessarily convert to m/s. your final answer of 51.1 m is slightly erroneous, not because of a mistake in the conversion, but because of premature rounding off. You say
s = 26.67*30 = 800.1 m
Actually it is 26.666666... * 30 = 800 m

The rule is always work to high precision and round off your final answer to the correct number of sig figs at the end. (If you are asked to report an intermediate result, report it to the correct sig figs, but still use the more precise value for subsequent calculations.)

Noted with Regards,

## 1. How do you calculate the distance between a car and a truck in mechanics?

The distance between a car and a truck can be calculated using the formula: Distance = Speed x Time. This formula takes into account the speed of the car and truck, as well as the time it takes for the car to reach the truck.

## 2. What factors affect the distance between a car and a truck?

The distance between a car and a truck can be affected by various factors such as the speed of the car and truck, the acceleration and deceleration of both vehicles, the road conditions, and the reaction time of the drivers.

## 3. Can the distance between a car and a truck change while they are in motion?

Yes, the distance between a car and a truck can change while they are in motion. This can happen if either vehicle changes its speed or direction, or if there are external factors such as other vehicles or obstacles on the road.

## 4. How can you use physics principles to find the distance between a car and a truck?

In physics, the distance between a car and a truck can be calculated using the principles of kinematics, which deals with the motion of objects. By considering the initial and final positions, velocities, and accelerations of both vehicles, the distance between them can be determined.

## 5. Is the distance between a car and a truck the same as the displacement?

No, the distance between a car and a truck is not the same as the displacement. Distance refers to the total length covered by both vehicles, while displacement refers to the shortest straight line distance between their initial and final positions.

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