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Physics A physics question: kinematics

leprofece

Member
Jan 23, 2014
241
A car is stopped at a traffic light in the moment in which a truck which has a constant speed of 50 Km/h
passes throught a trafic light light eats and passes the car 10.s later the traffic light changes and the car accelerates to 2 m/s2 for 15 sec then which, it continuous your travel at constant speed, when and where the car reach the truck?
answer. 22, 59s after the auto boot and 452, 65 m below the traffic light.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The first thing you want to do is convert the trucks constant speed to meters per second so that its units will be consistent with everything else. Something moving at $v$ kph can be converted to m/s as follows:

\(\displaystyle v\frac{\text{km}}{\text{hr}}\cdot\frac{1000\text{ m}}{1\text{ km}}\cdot\frac{1\text{ hr}}{3600\text{ s}}=\frac{5}{18}v\)

Now, to work the rest of the problem, I would use a graphical approach. The distance the truck moves can be represented by a rectangle whose base is the length of time it is moving (10 seconds more than the car) and whose height is the truck's speed.

The distance the car moves will consist of two shapes...a triangle for the time in which is is accelerating (base is 15) and whose height is the speed it is moving after 15 seconds, and then there is a rectangle for the time during which the car moves with constant speed. The sum of the two bases of the shapes associated with the car must be $t$. Can you proceed?

edit: I have moved this topic here as it is algebra based physics. :D
 

leprofece

Member
Jan 23, 2014
241
The first thing you want to do is convert the trucks constant speed to meters per second so that its units will be consistent with everything else. Something moving at $v$ kph can be converted to m/s as follows:

\(\displaystyle v\frac{\text{km}}{\text{hr}}\cdot\frac{1000\text{ m}}{1\text{ km}}\cdot\frac{1\text{ hr}}{3600\text{ s}}=\frac{5}{18}v\)

Now, to work the rest of the problem, I would use a graphical approach. The distance the truck moves can be represented by a rectangle whose base is the length of time it is moving (10 seconds more than the car) and whose height is the truck's speed.

The distance the car moves will consist of two shapes...a triangle for the time in which is is accelerating (base is 15) and whose height is the speed it is moving after 15 seconds, and then there is a rectangle for the time during which the car moves with constant speed. The sum of the two bases of the shapes associated with the car must be $t$. Can you proceed?
No could you please tell me the equations
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I choose to start the clock when the car begins moving, and the call the point in time where the care catches truck $t$, which is measured in seconds. So, the car will take $t$ seconds to catch the truck. How long will the truck have then been in motion since it passed the stationary car?
 

leprofece

Member
Jan 23, 2014
241
I choose to start the clock when the car begins moving, and the call the point in time where the care catches truck $t$, which is measured in seconds. So, the car will take $t$ seconds to catch the truck. How long will the truck have then been in motion since it passed the stationary car?
Ok I have truck time 10 + Tauto
Then Car distance is 1/2 at2 = 13,88(10 + Tauto)
but i Need to calculate more things To get the answer
this is the style I need
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, the distance the car moves in meters during acceleration is:

\(\displaystyle d_{C_1}=\frac{1}{2}(2)t^2=t^2\)

It accelerates for 15 seconds, hence:

\(\displaystyle d_{C_1}=15^2=225\)

Now, what is the speed of the car once it stops accelerating? Recall that for constant acceleration and an initial speed of zero, we have:

\(\displaystyle v=at\)

So what is the car's speed after 15 seconds of accelerating at 2 m/s²?

And then use distance equals speed times time to compute how far the car moves at this constant speed, and keep in mind that the sum of the time spent accelerating (15 seconds) and the time spent at constant velocity must be $t$ seconds. So, what is the time spent at constant velocity?
 

leprofece

Member
Jan 23, 2014
241
must I equal what??
with this point of view You get 15 that is nopt the answer
the quadratic equation is t^2 +138,88+13,88 t= 0???
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
must I equal what??
with this point of view You get 15 that is nopt the answer
the quadratic equation is t^2 +138,88+13,88 t= 0???
I don't know how you are arriving at that quadratic. After 15 seconds of acceleration, the car is moving at 30 m/s:

\(\displaystyle v=at=2(15) = 30\)

Now, since it has already moved for 15 seconds, it will move for $t-15$ seconds at this speed, so its total distance moved is:

\(\displaystyle d_{C}=225+30(t-15)=30t-225\)

Now, you want to compute the distance moved by the truck during its $t+10$ seconds of movement, and then equate the two distances and solve for $t$. What is the speed of the truck in meters per second? Once you have this, then multiply it times the duration of time I gave above to get the distance moved by the truck. What do you find?
 

leprofece

Member
Jan 23, 2014
241
I don't know how you are arriving at that quadratic. After 15 seconds of acceleration, the car is moving at 30 m/s:

\(\displaystyle v=at=2(15) = 30\)

Now, since it has already moved for 15 seconds, it will move for $t-15$ seconds at this speed, so its total distance moved is:

\(\displaystyle d_{C}=225+30(t-15)=30t-225\)

Now, you want to compute the distance moved by the truck during its $t+10$ seconds of movement, and then equate the two distances and solve for $t$. What is the speed of the truck in meters per second? Once you have this, then multiply it times the duration of time I gave above to get the distance moved by the truck. What do you find?
What two?? I see only one
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
What two?? I see only one
You should get an expression for the distance moved by the truck and an expression for the distance moved by the car, both of which involve $t$. Since these distances must be equal, we can equate them and solve for $t$.
 

leprofece

Member
Jan 23, 2014
241
BUT which are these two equations joined in one ???
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
BUT which are these two equations joined in one ???
I gave the the expression for the car's distance. You need to find the expression for the truck's distance, and since the truck's speed is constant, it is just the product of the speed and the duration of time it is moving (which I gave). Be sure to convert the truck's speed to meters per second...I showed you how to do this.

You should now be able to put the pieces together...what do you find for the truck's distance as a function of time $t$?