Speed of sound - Uniform motion- linear equation application

[paulmdrdo1]

I need help with these problems.

1. Speed of Sound in Air. Two seconds after firing a rifle at a
target, the shooter hears the impact of the bullet. Sound travels
at 1100 feet per second and the bullet at 1865 feet per second.
Determine the distance to the target (to the nearest foot).

2. Uniform Motion. A car traveling at 80 kilometers per hour is
passed by a second car going in the same direction at a constant
speed. After 30 seconds, the two cars are 500 meters apart.
Find the speed of the second car.

my attempt for 1st.

1100(t) = distance traveled by sound
1865(t+2)= distance traveled by bullet

1865(t+2)=1100(t)

but i get a negative answer. this is where I can get to.

my attempt for 2

let r = the speed of the second car.

1/45 kps = rate of the first car

30r-2/3=500

r=16.69 kps.

please help me with these. thanks!

 

[MarkFL]

For the first problem, let's let $t_1$ be the time in seconds it takes the bullet to reach the target:

\(\displaystyle d=1865t_1\)

And let $t_2$ be the time it takes for the sound of the bullet striking the target to reach the shooter:

\(\displaystyle d=1100t_2\)

We are given that:

\(\displaystyle t_1+t_2=2\)

You now have 3 unknowns and 3 equations. We can use the last equation to eliminate $t_2$ from the second equation and then have two equations in two unknowns:

\(\displaystyle d=1865t_1\)

\(\displaystyle d=1100\left(2-t_1 \right)\)

I would next use the first equation to eliminate $t_1$ from the second equation:

\(\displaystyle d=1100\left(2-\frac{d}{1865} \right)\)

Now, you can solve for $d$. What do you find?

 

[paulmdrdo1]

so the total time taken for the bullet to reach the target and for the sound to reach the shooter is 2 secs? can you explain. why that is?

 

[MarkFL]

so the total time taken for the bullet to reach the target and for the sound to reach the shooter is 2 secs? can you explain. why that is?

This is what the problem states:

" Two seconds after firing a rifle at a target, the shooter hears the impact of the bullet."

 

[paulmdrdo1]

this my attempt using one variable only

1100t=1865(2-t)

1100t=3730-1865t
2965t=3730

t = 1 and 153/593

the distance of the target from the shooter is 1383.8 ft or 1384 ft.

is this correct?

 

[MarkFL]

I get 1384 ft. (to the nearest foot) as well.

 

[paulmdrdo1]

how about the second problem?

 

[MarkFL]

Let's let the speed of the faster car be relative to the slower car...how far did the faster car travel (in kilometers) relative to the slower car and in what amount of time (in hours)? Use this in the equation:

\(\displaystyle d=vt\)

Then solve for $v$ and add this value to the speed of the slower car...what do you find?

 

[paulmdrdo1]

MarkFL, can you show the method using just one variable. :D

what do you mean by this "let the speed of the faster car be relative to the slower car."?

 

[MarkFL]

MarkFL, can you show the method using just one variable. :D

what do you mean by this "let the speed of the faster car be relative to the slower car."?

You know $d$ and $t$, so the only variable left is that which you seek: $v$.

In our coordinate system, $v$ represents the speed of the faster car in excess of that of the slower car. Once we find $v$, then we will add it to the speed of the slower car to get the faster car's actual speed relative to the surroundings.

 

[paulmdrdo1]

let r = speed of the second car

what I have in mind is the two cars are at the same distance when they passed each other
then after 30 secs they are 500 meters apart, meaning the second car has traveled 500 meters away from first car from their same distance position.

if the car traveled 500 meters in 30 sec. its speed r = 50/3 meters/sec

I'll add it to the speed of the slower car so that I can get the speed of the faster car

80kph = 200/9 meters/sec

200/9 +50/3= 350/9 meters/sec

350/9 meters/sec = 140 kph --->> speed of the faster car.

is my thought process here correct? thanks!

 

[MarkFL]

Yes, that's the same result I obtained. I let $v_R$ be the speed of the faster car relative to the slower car and stated:

\(\displaystyle \frac{1}{2}\text{ km}=v_R\frac{1}{120}\text{ hr}\)

\(\displaystyle v_R=60\,\frac{\text{km}}{\text{hr}}\)

Add this to the speed of the slower car to get the absolute speed $v$ of the faster car:

\(\displaystyle v=\left(80+60 \right)\, \frac{\text{km}}{\text{hr}}= 140\,\frac{\text{km}}{\text{hr}}\)