How long did it take for the bullet to hit the target & what is the distance of the target in meters in feet?

Question

This is from Chapter 5, Exercise 5-3, #16 of Fundamentals of Technical Mathematics (2nd edition): Police officer Sabrena fires a rifle at a practice target and hears the bullet strike 1.5 seconds later. The muzzle velocity of the rifle is 1000 meter per second (3280 feet per second), and the speed of sound is 332 meters per second (1090 feet per second). If both velocities are constant, how long did it take for the bullet to hit the target, and what is the distance of the target in meters and in feet?

Answer 1

Interesting question....

The 1.5 seconds really has two components...the time it takes for the bullet to strike the target and the time it takes for the sound to reach the officer from the target.

I apologize that I am leaving some of the units off the numbers. It gets too confusing if all the units and such are left on when typing in a forum like this.

First, we will work with meters. I will use the variables "X" and "Y" for the time variables since you really can't use subscript on Yahoo...

1) 1000X =D
2) 332Y = D
3) X + Y = 1.5

Solving eq 3 for "y" Y = 1.5 - X

Substituting eq 3 into eq 2

332(1.5 -X) = D = 498 - 332X

Substituting the value of "D" in the previous step into eq 1

1000X = 498 - 332X
1332X = 498
X = 0.374 seconds (rounded)

D = 1000X = 1000 (.374) = 374 meters

Check your work using the speed of sound:

0.374 + Y = 1.5
Y = 1.126 sec

D = 332Y
D = 332(1.126)
D= 373.8 meters (slight rounding error)

Now that you know the time, you can adjust the equation slightly to find the number of feet to the target:

D = 3280X
X = 0.374 seconds

D = 1227 feet